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THE  CATHERINE  B.  O'CONNOR  LIBRARV 

WESTON  OBSERVATORY 

WESTON,  MASSACHUSETTS    02193 


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FIRST    COURSE 

IN  THE 

THEORY    OF    EQUATIONS 


FIRST  COURSE 

IN  THE 

THEORY   OF  EQUATIONS 

QAM 

LEONARD  EUGENE  DICKSON,  Ph.D. 


CORRESPONDANT    DE    L  INSTITUT    DE    FRANCE 
PROFESSOR   OF   MATHEMATICS   IN   THE   UNIVERSITY   OF   CHICAGO 


1??1 


LIBRARY  SEROLOGICAL 
OBSERVATORY 


"*''*■'"'-•* 


NEW  YORK 

JOHN  WILEY  &  SONS,  Inc. 

London:  CHAPMAN  &  HALL,  Limited 

THE  CATHERINE  B.  O'CONNOR  LIBRARY 

WESTON  OBSERVATORY 

WESTON,  MASSACHUSETTS    02193 


499 


Copyright,  1922, 

BY 

LEONARD  EUGENE  DICKSON 


All  Rights  Reserved 
This  book  or  any  part  thereof  must  not 
be   reproduced   in   any  form   without 
the  written  permission  of  the  publisher. 


.Printed  in  U.  S.  A 

5/32 


PRESS    OF 

BR/.UNWORTH    &.    CO.,    INC. 

BOOK    MANUFACTURERS 

BROOKLYN,   NEW  YORK 


THE  CATHERINE  B.  O'CONNOR  LIBRARY 
WESTON  OBSERVATORY 

WESTON,  MASSACHUSETTS    02193 


PREFACE 


The  theory  of  equations  is  not  only  a  necessity  in  the  subsequent 
mathematical  courses  and  their  applications,  but  furnishes  an  illumina- 
ting sequel  to  geometry,  algebra  and  analytic  geometry.  Moreover, 
it  develops  anew  and  in  greater  detail  various  fundamental  ideas  of 
calculus  for  the  simple,  but  important,  case  of  polynomials.  The 
theory  of  equations  therefore  affords  a  useful  supplement  to  differential 
calculus  whether  taken  subsequently  or  simultaneously. 

It  was  to  meet  the  numerous  needs  of  the  student  in  regard  to  his 
earlier  and  future  mathematical  courses  that  the  present  book  was 
planned  with  great  care  and  after  wide  consultation.  It  differs  essentially 
from  the  author's  Elementary  Theory  of  Equations,  both  in  regard  to 
omissions  and  additions,  and  since  it  is  addressed  to  younger  students 
and  may  be  used  parallel  with  a  course  in  differential  calculus.  Simpler 
and  more  detailed  proofs  are  now  employed.  The  exercises  are  simpler, 
more  numerous,  of  greater  variety,  and  involve  more  practical  applications. 

This  book  throws  important  light  on  various  elementary  topics. 
For  example,  an  alert  student  of  geometry  who  has  learned  how  to  bisect 
any  angle  is  apt  to  ask  if  every  angle  can  be  trisected  with  ruler  and 
compasses  and  if  not,  why  not.  After  learning  how  to  construct  regular 
polygons  of  3,  4,  5,  6,  8  and  10  sides,  he  will  be  inquisitive  about  the 
missing  ones  of  7  and  9  sides.  The  teacher  will  be  in  a  comfortable  position 
if  he  knows  the  facts  and  what  is  involved  in  the  simplest  discussion  to 
date  of  these  questions,  as  given  in  Chapter  III.  Other  chapters  throw 
needed  light  on  various  topics  of  algebra.  In  particular,  the  theory 
of  graphs  is  presented  in  Chapter  V  in  a  more  scientific  and  practical 
manner  than  was  possible  in  algebra  and  analytic  geometry. 

There  is  developed  a  method  of  computing  a  real  root  of  an  equation 
with  minimum  labor  and  with  certainty  as  to  the  accuracy  of  all  the 
decimals  obtained.     We  first  find  by  Horner's  method  successive  trans- 


iv  PREFACE 

formed  equations  whose  number  is  half  of  the  desired  number  of  significant 
figures  of  the  root.  The  final  equation  is  reduced  to  a  linear  equation 
by  applying  to  the  constant  term  the  correction  computed  from  the 
omitted  terms  of  the  second  and  higher  degrees,  and  the  work  is  completed 
bjr  abridged  division.  The  method  combines  speed  with  control  of 
accuracy. 

Newton's  method,  which  is  presented  from  both  the  graphical  and 
the  numerical  standpoints,  has  the  advantage  of  being  applicable  also  to 
equations  which  are  not  algebraic;  it  is  applied  in  detail  to  various  such 
equations. 

In  order  to  locate  or  isolate  the  real  roots  of  an  equation  we  may 
employ  a  graph,  provided  it  be  constructed  scientifically,  or  the  theorems 
of  Descartes,  Sturm,  and  Budan,  which  are  usually  neither  stated,  nor 
proved,  correctly. 

The  long  chapter  on  determinants  is  independent  of  the  earlier  chap- 
ters. The  theory  of  a  general  system  of  linear  equations  is  here  pre- 
sented also  from  the  standpoint  of  matrices. 

For  valuable  suggestions  made  after  reading  the  preliminary  manu- 
script of  this  book,  the  author  is  greatly  indebted  to  Professor  Bussey 
of  the  University  of  Minnesota,  Professor  Roever  of  Washington  Uni- 
versity, Professor  Kempner  of  the  University  of  Illinois,  and  Professor 
Young  of  the  University  of  Chicago.  The  revised  manuscript  was  much 
improved  after  it  was  read  critically  by  Professor  Curtiss  of  Northwestern 
University.  The  author's  thanks  are  due  also  to  Professor  Dresden  of 
the  University  of  Wisconsin  for  various  useful  suggestions  on  the 
proof-sheets. 
Chicago,  1921. 


CONTENTS 

Numbers  refer  to  pages. 


i 


CHAPTER  I 

Complex  Numbers 
Square  roots,  1.     Addition,  multiplication  and  division    of  complex  num- 
bers,   2.     Cube  roots  of  unity,  3.     Geometrical  representation,    3.     Product 
and  quotient,  4.     De  Moivre's  theorem,   5.     Cube  roots,    5.     nth  roots,  7. 
Roots  of  unity,  8,     Primitive  roots  of  unity,  9. 

CHAPTER  II 

Elementary  Theorems  on  the  Roots  of  an  Equation 
Quadratic  equation,  11.  Remainder  theorem,  12.  Synthetic  division,  13. 
Factored  form  of  a  polynomial,  15.  Multiple  roots,  16.  Identical  poly- 
nomials, 16.  Relations  between  the  roots  and  the  coefficients,  17.  Imaginary 
roots  occur  in  pairs,  19.  Upper  limit  to  the  real  roots,  21.  Integral  roots, 
24.     Rational  roots,  27. 

CHAPTER   III 

Constructions  with  Ruler  and  Compasses 
Graphical  solution  of  a   quadratic  equation,  29.      Analytic   criterion    for 
constructibility,   30.     Cubic  equations  with  a  constructible  root,   32.     Tri- 
section  of  an  angle,  34.     Duplication  of  a  cube,  35.     Regular  polygons  of 
7,  9,  17,  and  n  sides,  35-44.     Reciprocal  equations,  37. 

CHAPTER  IV 

Cubic  and  Quartic  Equations 

Algebraic  solution  of  a  cubic,  45.     Discriminant,  47.     Number  of  real  roots 

of  a  cubic,  48.     Trigonometric  solution  of  a  cubic,  49.      Ferrari's  and  Descartes' 

solutions  of  a  quartic,  50.     Resolvent  cubic,  51 .     Discriminant  of  a  quartic,  51 . 

CHAPTER   V 
The  Graph  of  an  Equation 
Use  of  graphs,  55.     Caution  in  plotting,  55.     Bend  points,  56.     Derivatives, 
58.     Horizontal  tangents,  60.     Multiple  roots,  60.     Ordinary  and  inflexion 

v 


vi  CONTENTS 

tangents,  62.  Real  roots  of  a  cubic  equation,  65.  Continuity,  66.  Con- 
dition for  a  root  between  a  and  b,  67.  Sign  of  a  polynomial  at  infinity,  68. 
Rolle's  theorem,  69. 

CHAPTER  VI 
Isolation  of  the  Real  Roots 
Descartes'  rule  of  signs,  71.     Sturm's  method,  75.     Sturm's  functions  for 
the  general  quartic  equation,  80.     Budan's  theorem,  83. 

CHAPTER  VII 

Solution  of  Numerical  Equations 
Horner's  method,  86.     Newton's  method,  algebraic  and  graphical  discus- 
sion, systematic  computation,  also  for  functions  not  polynomials,  90.     Imagi- 
nary roots,  98. 

CHAPTER  VIII 

Determinants;  Systems  of  Linear  Equations 
Solution  of  2  or  3  linear  equations  by  determinants,  101.  Even  and  odd 
arrangements,  103.  Definition  of  a  determinant  of  order  n,  105.  Interchange 
of  rows  and  columns,  106.  Interchange  of  two  columns  or  two  rows,  107. 
Minors,  109.  Expansion,  109.  Removal  of  factors,  111.  Sum  of  deter- 
minants, 112.  Addition  of  columns  or  rows,  113.  Rank,  116.  System 
of  n  linear  equations  in  n  unknowns,  114,  116.  Homogeneous  equations,  119. 
System  of  m  linear  equations  in  n  unknowns,  matrix  and  augmented  matrix, 
120.  Complementary  minors,  122.  Laplace's  development,  122.  Product  of 
determinants,  124. 

CHAPTER  IX 

Symmetric  Functions 
Sigma   functions,    128.     Elementary    symmetric   functions,    128.     Funda- 
mental theorem,  129.     Rational  functions  symmetric  in  all  but  one  of  the 
roots,    132.     Sums  of  like  powers  of  the  roots,   Newton's  identities,    134. 
Waring's  formula,  136.     Computation  of  symmetric  functions,  141. 

CHAPTER  X 

Elimination,  Resultants  and  Discriminants 
Methods  of  Sylvester,  Euler,  and  Bezout,  143.     Discriminants,  152. 

APPENDIX 

The  Fundamental  Theorem  of  Algebra 

Answers 159 

Index 167 


First  Course  in 
The  Theory  of  Equations 


CHAPTER  I 

Complex  Numbers 

1.  Square  Roots.  If  p  is  a  positive  real  number,  the  symbol  Vp  is 
used  to  denote  the  positive  square  root  of  p.  It  is  most  easily  computed 
by  logarithms. 

We  shall  express  the  square  roots  of  negative  numbers  in  terms  of  the 
symbol  i  such  that  the  relation  i2  =  —  1  holds.  Consequently  we  denote 
the  roots  of  x2  =  —  1  by  i  and  —i.  The  roots  of  x2  =  —  4  are  written  in  the 
form  ±2i  in  preference  to  ±  V  —  4.  In  general,  if  p  is  positive,  the  roots 
of  x2—  —  p  are  written  in  the  form  ±Vp  i  in  preference  to  ±  v  —  p. 

The  square  of  either  root  is  thus  (V p)H2  —  —  p.  Had  we  used  the  less  desirable 
notation  ±  v  —  P  for  the  roots  of  x2=  —  p,  we  might  be  tempted  to  find  the  square  of 
either  root  by  multiplying  together  the  values  under  the  radical  sign  and  conclude 
erroneously  that 

V-p  V-p  =  Vp2=+p. 

To  prevent  such  errors  we  use  V  p  i  and  not  V  —p. 

2.  Complex  Numbers.  If  a  and  6  are  any  two  real  numbers  and 
4?=—l,a-\-bi  is  called  a  complex  number  1  and  a  —  6z  its  conjugate.  Either 
is  said  to  be  zero  if  a  =  b  =  0.  Two  complex  numbers  a-\-bi  and  c+d« 
are  said  to  be  equal  if  and  only  if  a  =  c  and  b  =  d.     In  particular,  a+fo'  =  0 

1  Complex  numbers  are  essentially  couples  of  real  numbers.  For  a  treatment  from 
this  standpoint  and  a  treatment  based  upon  vectors,  see  the  author's  Elementary  Theory 
oj  Equations,  p.  21,  p.  18. 


2  COMPLEX  NUMBERS  [Ch.  I 

if  and  only  if  a  =  b  =  0.     If  6^0,  a-\-bi  is  said  to  be  imaginary.     In  partic- 
ular, bi  is  called  a  pure  imaginary. 

Addition  of  complex  numbers  is  defined  by 

(a+bi)  +  {c+di)  =  (a+c)  +  (b+d)i. 

The  inverse  operation  to  addition  is  called  subtraction,  and  consists  in 
finding  a  complex  number  z  such  that 

(c-\-di)-\-z  =  a-\-bi. 
In  notation  and  value,  z  is 

(a-\-bi)  -  {c+di)  =  {a-c)  +  {b-d)i. 

Multiplication  is  defined  by 

(a-\-bi)  (c-\-di)  —ac  —  bd-\-{ad-{-bc)i, 

and  hence  is  performed  as  in  formal  algebra  with  a  subsequent  reduction 
by  means  of  i2  =  —  1.     For  example, 

(a+bi)  (a-bi)=a2-b2i2  =  a2+b2. 

Division  is  defined  as  the  operation  which  is  inverse  to  multiplication, 
and  consists  in  finding  a  complex  number  q  such  that  (a+bi)q  =  e+fi. 
Multiplying  each  member  by  a  —  bi,  we   find  that  q  is,  in  notation  and 

value, 

e  -\-ji  _  (e  +fi)  {a  —  bi)  _  ae + bf  ,    af—be. 


a+bi  a2+b2  a2+b2  '    a2-\-b2 

Since  a2+62  =  0  implies  a  =  b  =  0  when  a  and  b  are  real,  we  conclude  that 
division  except  by  zero  is  possible  and  unique. 

EXERCISES 

Express  as  complex  numbers 

1.  V^9.  2.  VI.  3.  (V25+V-25)  V^16-      4.   -f . 

/-  3+V-5  „    3+5i  a+bi 

5.  8+2V3.  6.  ==.  7.  — - .  8.  — - . 

2-|-V  —  1  2— 6%  a— 01 

9.  Prove  that  the  sum  of  two  conjugate  complex  numbers  is  real  and  that  their 
difference  is  a  pure  imaginary. 

10.  Prove  that  the  conjugate  of  the  sum  of  two  complex  numbers  is  equal  to  the 
sum  of  their  conjugates.  Does  the  result  hold  true  if  each  word  sum  is  replaced  by  the 
word  difference? 


§4] 


GEOMETRICAL  REPRESENTATION 


11.  Prove  that  the  conjugate  of  the  product  (or  quotient)  of  two  complex  numbers 
is  equal  to  the  product  (or  quotient)  of  their  conjugates. 

12.  Prove  that,  if  the  product  of  two  complex  numbers  is  zero,  at  least  one  of  them 
is  zero. 

13.  Find  two  pairs  of  real  numbers  x,  y  for  which 

(x+yi)2=  -7+24t. 
As  in  Ex.  13,  express  as  complex  numbers  the  square  roots  of 

14.    -11+60  i.  15.5-121  16.  ±cd+(2c2-2d2)i. 

3.  Cube  Roots  of  Unity.  Any  complex  number  x  whose  cube  is  equal 
to  unity  is  called  a  cube  root  of  unity.     Since 

x3-l  =  (x-l)  (x2+x+l), 
the  roots  of  x3  =  1  are  1  and  the  two  numbers  x  for  which 

x2+x+l  =  0,        (z+i)2=-f,        z+i=±aV3z\ 
Hence  there  are  three  cube  roots  of  unity,  viz., 

1,         a=-$+$VZi,         a/=-!-iV3;. 
In  view  of  the  origin  of  a>,  we  have  the  important  relations 

W2  +  CO+1=0,       co3  =  l. 

Since  coa/  =  l  and  co3  =  l,  it  follows  that  a/  =  co2,  co  =  w'2. 

4.  Geometrical  Representation  of  Complex  Numbers.  Using  rect- 
angular axes  of  coordinates,  OX  and  OY,  we  represent  the  complex  number 
a-\-bi  by  the  point  A  having  the  coordinates  a,  b  (Fig.  1). 

The  positive  number  r=Va2+b2  giving 
the  length  of  OA  is  called  the  modulus  (or 
absolute  value)  of  a -\-bi.  The  angle  6  =  XOA, 
measured  counter-clockwise  from  OX  to  OA, 
is  called  the  amplitude  (or  argument)  of  a-{-bi. 
Thus  cos  6  =  a/r,  sin  6  =  b/r,  whence 

(1)  a+fo'  =  r(cos  0-H  sin  6). 


Fig.  1 


The  second  member  is  called  the  trigonometric  form  of  a-\-bi. 

For  the  amplitude  we  may  select,  instead  of  6,  any  of  the  angles  0±36O°, 
0±72O°.  etc. 


COMPLEX  NUMBERS 


[Ch.  I 


Fig.  2 


Two  complex  numbers  are  equal  if  and  only  if  their  moduli  are  equal 
and  an  amplitude  of  the  one  is  equal  to  an  amplitude  of  the  other. 

For  example,  the  cube  roots  of  unity  are  1  and 

=  cos  120°  +i  sin  120°, 
co2=-i-iV3  i 

=  cos240°+isin240°, 

and  are  represented  by  the  points  marked  1,  u,  a2 
at  the  vertices  of  an  equilateral  triangle  inscribed 
in  a  circle  of  radius  unity  and  center  at  the  origin 
O  (Fig.  2).  The  indicated  amplitudes  of  a>  and  w2 
are  120°  and  240°  respectively,  while  the  modulus 
of  each  is  1. 

The  modulus  of  —3  is  3  and  its  amplitude  is  180°  or  180°  plus  or  minus  the  product 
of  360°  by  any  positive  whole  number. 

5.  Product  of  Complex  Numbers.     By  actual  multiplication, 

[r(cos  d-\-i  sin  6)}  [r'(cos  a-\-i  sin  a)] 
=  rr'[(cos  6  cos  a  — sin  6  sina)-H(sin  6  cos  a+cos  6  sin  a)] 
=  rr'[cos  {6-\-a)-\-i  sin  (0+a)],      by  trigonometry. 

Hence  the  modulus  of  the  product  of  two  complex  numbers  is  equal  to  the 
product  of  their  moduli,  while  the  amplitude  of  the  product  is  equal  to  the 
sum  of  their  amplitudes. 

For  example,  the  square  of  w  =  cos  120°+£  sin  120°  has  the  modulus  1  and  the  ampli- 
tude 120°+120°  and  hence  is  «2  =  cos  240°+i  sin  240°.  Again,  the  product  of  w  and  co2 
has  the  modulus  1  and  the  amplitude  120°+240°  and  hence  is  cos  360°+*  sin  360°, 
which  reduces  to  1.     This  agrees  with  the  known  fact  that  w3  =  l. 

Taking  r  =  r'  =  1  in  the  above  relation,  we  obtain  the  useful  formula 
(2)  (cos  d+i  sin  6)  (cos  a+i  sin  a)  =cos  (d-\-a)+i  sin  (d+a). 

6.  Quotient  of  Complex  Numbers.  Taking  a  =  /3  —  6  in  (2)  and  divid- 
ing the  members  of  the  resulting  equation  by  cos  6-\-i  sin  6,  we  get 

cos  /3+i  sin  /3  .  '      x  .  .   .     ,_  ', 

.  .  .    .    ^  =  cos  G3-0)+* sin  03-0). 

cos  d-\-i  sin  6 


§  7]  DE  MOIVRE'S  THEOREM  5 

Hence  the  amplitude  of  the  quotient  of  R(cos  /3+t  sin  (3)  by  r (cos  d-\-i  sin  0) 
zs  egitaZ  to  the  difference  (3— 9  of  their  amplitudes,  while  the  modulus  of  the 
quotient  is  equal  to  the  quotient  R/r  of  their  moduli. 
The  case  /3  =  0  gives  the  useful  formula 

,  .   . — ^  =  cos  6  —  i  sin  6. 

cos  d-\-i  sin  6 

7.  De  Moivre's  Theorem.     //  n  is  any  positive  whole  number, 

(3)  (cos  d-\-i  sin  0)ra  =  cos  nd-\-i  sin  nd. 

This  relation  is  evidently  true  when  n  =  l,  and  when  n  =  2  it  follows 
from  formula  (2)  with  a  =  6.  To  proceed  by  mathematical  induction, 
suppose  that  our  relation  has  been  established  for  the  values  1,2,  ...  ,m 
of  n.  We  can  then  prove  that  it  holds  also  for  the  next  value  ra+1  of  n. 
For,  by  hypothesis,  we  have 

(cos  0-\-i  sin  0)OT  =  cos  md-\-i  sin  md. 

Multiply  each  member  by  cos  0-\-i  sin  6,  and  for  the  product  on  the  right 
substitute  its  value  from  (2)  with  a  =  md.     Thus 

(cos  6-\-i  sin  d)m+1  =  (cos  6-\-i  sin  6)  (cos  md-\-i  sin  md), 

=  cos  (6-\-md)-\-ism  (d-\-md), 

which  proves  (3)  when  w  =  m+l.     Hence  the  induction  is  complete. 

Examples  are  furnished  by  the  results  at  the  end  of  §  5 : 

(cos  120°  +i  sin  120°)2  =  cos  240°  +i  sin  240°, 
(cos  120°+i  sin  120°)3  =  cos  360°+i  sin  360°. 

8.  Cube  Roots.  To  find  the  cube  roots  of  a  complex  number,  we  first 
express  the  number  in  its  trigonometric  form.     For  example, 

4V2+4V2  i  =  8(cos  45°+t  sin  45°). 

If  it  has  a  cube  root  which  is  a  complex  number,  the  latter  is  expressible 
in  the  trigonometric  form 

(4)  r(cos  d-\-i  sin  6). 

The  cube  of  the  latter,  which  is  found  by  means  of  (3),  must  be  equal 
to  the  proposed  number,  so  that 

r3(cos  Z6+i  sin  30)  =8(cos  45°+t  sin  45°). 


6 


COMPLEX  NUMBERS 


[Ch.  I 


The  moduli  r3  and  8  must  be  equal,  so  that  the  positive  real  number  r 
is  equal  to  2.  Furthermore,  30  and  45°  have  equal  cosines  and  equal 
sines,  and  hence  differ  by  an  integral  mulitple  of  360°.  Hence  30  =  45°+ 
k  -360°,  or  0=15°+fc-12O°,  where  k  is  an  integer.1  Substituting  this 
value  of  0  and  the  value  2  of  r  in  (4),  we  get  the  desired  cube  roots.  The 
values  0,  1,  2  of  k  give  the  distinct  results 

#i=2(cos  15°+*' sin  15°),  i?2  =  2(cos  135°+i  sin  135°), 

#3  =  2(cos255°+;sin2550). 

Each  new  integral  value  of  k  leads  to  a  result  which  is  equal  to  Ri, 
R2  or  R3.  In  fact,  from  A;  =  3  we  obtain  R\,  from  fc  =  4  we  obtain  R2,  from 
k  =  5  we  obtain  R3,  from  /b  =  6  we  obtain  Ri  again,  and  so  on  periodically. 

EXERCISES 

1.  Verify  that  R2  =  uRi,  Ri  =  ui'lRi.  Verify  that  Ri  is  a  cube  root  of  8  (cos  45°  + 
i  sin  45°)  by  cubing  Ri  and  applying  De  Moivre's  theorem.  Why  are  the  new  expressions 
for  R2  and  R3  evidently  also  cube  roots? 

2.  Find  the  three  cube  roots  of  —27;  those  of  —  i;  those  of  u. 

3.  Find  the  two  square  roots  of  i;  those  of  —  i;  those  of  w. 

4.  Prove  that  the  numbers  cos  d+i  sin  0  and  no  others  are  represented  by  points 
on  the  circle  of  radius  unity  whose  center  is  the  origin. 

5.  If  a+bi  and  c+di  are  represented  by  the  points  A  and  C  in  Fig.  3,  prove  that 
their  sum  is  represented  by  the  fourth  vertex  S  of  the  parallelogram  two  of  whose  sides 
are  OA  and  OC.  Hence  show  that  the  modulus  of  the  sum  of  two  complex  numbers 
is  equal  to  or  less  than  the  sum  of  their  moduli,  and  is  equal  to  or  greater  than  the  dif- 
ference of  their  moduli. 


1  Here,  as  elsewhere  when  the  contrary  is  not  specified,  zero  and  negative  as  well 
as  positive  whole  numbers  are  included  under  the  term  "integer." 


§  9]  ROOTS  OF  COMPLEX  NUMBERS  7 

6.  Let  r  and  r'  be  the  moduli  and  6  and  a  the  amplitudes  of  two  complex  numbers 
represented  by  the  points  A  and  C  in  Fig.  4.  Let  U  be  the  point  on  the  x-axis  one 
unit  to  the  right  of  the  origin  O.  Construct  triangle  OCP  similar  to  triangle  OUA  and 
similarly  placed,  so  that  corresponding  sides  are  OC  and  OU,  CP  and  UA,  OP  and  OA, 
while  the  vertices  0,  C,  P  are  in  the  same  order  (clockwise  or  counter-clockwise)  as 
the  corresponding  vertices  O,  U,  A.  Prove  that  P  represents  the  product  ( §  5)  of  the 
complex  numbers  represented  by  A  and  C. 

7.  If  a+bi  and  e+fi  are  represented  by  the  points  A  and  S  in  Fig.  3,  prove  that 
the  complex  number  obtained  by  subtracting  a-\-bi  from  e-\-fi  is  represented  by  the  point 
C.  Hence  show  that  the  absolute  value  of  the  difference  of  two  complex  numbers  is 
equal  to  or  less  than  the  sum  of  their  absolute  values,  and  is  equal  to  or  greater  than 
the  difference  of  their  absolute  values. 

8.  By  modifying  Ex.  6,  show  how  to  construct  geometrically  the  quotient  of  two 
complex  numbers, 

9.  nth  Roots.  As  illustrated  in  §  8,  it  is  evident  that  the  nth  roots 
of  any  complex  number  p (cos  A-\- i  sin  A)  a^e  the  products  of  the  nth 
roots  of  cos  A  -\-i ^  sin  A  by  the  positive  real  nth  root  of  the  positive  real 
number  p  (which  may  be  found  by  logarithms). 

Let  an  nth  root  of  cos  A-\-i  sin  A  be  of  the  form 

(4)  r(cos  6+i  sin  &). 
Then,  by  De  Moivre's  theorem, 

rn(cos  nd-\-i  sin  nd)  =  cos  A  -\-i  sin  A. 

The  moduli  rn  and  1  must  be  equal,  so  that  the  positive  real  number  r 
is  equal  to  1.  Since  nd  and  A  have  equal  sines  and  equal  cosines,  they 
differ  by  an  integral  multiple  of  360°.  Hence  nd  =  A-\-k-3Q0°,  where  k 
is  an  integer.  Substituting  the  resulting  value  of  6  and  the  value  1  of  r 
in  (4),  we  get 

(5)  cos  I )  +i  sin 


For  each  integral  value  of  k,  (5)  is  an  answer  since  its  nth  power  reduces 
to  cos  A-\-i  sin  A  by  DeMoivre's  theorem.  Next,  the  value  n  of  k  gives 
the  same  answer  as  the  value  0  of  k;  the  value  n+1  of  k  gives  the  same 
answer  as  the  value  1  of  A;;  and  in  general  the  value  n+m  of  k  gives  the 
same  answer  as  the  value  m  of  i  Hence  we  may  restrict  attention  to 
the  values  0,  1,  .  .  .  ,  n— 1  of  k.     Finally,  the  answers  (5)  given  by  these 


Observatory 
Weston  College 
Weston  Massachusetts 


8 


COMPLEX  NUMBERS 


[Ch.  I 


values  0,  1,  .  .  .  ,  n—  1  of  k  are  all  distinct,  since  they  are  represented  by 
points  whose  distance  from  the  origin  is  the  modulus  1  and  whose  ampli- 
tudes are 

A  A    360^  A    2-360° 

i  «  ~t~  }  • 

n  n 


A    360c 
n        n 


A     (n-l)360e 

n  n 


so  that  these  n  points  are  equally  spaced  points  on  a  circle  of  radius  unity. 
Special  cases  are  noted  at  the  end  of  §  10.  Hence  any  complex  number 
different  from  zero  has  exactly  n  distinct  complex  nth  roots. 

10.  Roots  of  Unity.     The  ti'igonometric  form  of  1  is  cos  0°-H  sin  0°. 
Hence  by  §  9  with  A  =  0,  the  n  distinct  nth  roots  of  unity  are 


(6) 


2/C7T         .      .       2/C7T        n  . 

cos \-i  sin (/c  =  0,  1,  .  .  .  ,  n— 1), 

n  n 


where  now  the  angles  are  measured  in  radians  (an  angle  of  180  degrees 
being  equal  to  -k  radians,  where  tt  =  3.1416,  approximately).  For  k  =  0, 
(6)  reduces  to  1,  which  is  an  evident  nth.  root  of  unity.     For  k=l,  (6)  is 


(7) 


it  =  cos \-i  sin  — . 

n  n 


By  De  Moivre's  theorem,  the  general  number  (6)  is  equal  to  the 
fcth  power  of  R.    Hence  the  n  distinct  nth  roots  of  unity  are 

(8)  R,R2,R3,  .  .  .  ,Rn-\Rn=h 

As  a  special  case  of  the  final  remark  in  §  9,  the  n  complex  numbers 
(6),  and  therefore  the  numbers  (8),  are  represented  geometrically  by  the 
vertices  of  a  regular  polygon  of  n  sides  inscribed  in  the  circle  of  radius 
unity  and  center  at  the  origin  with  one  vertex  on  the  positive  rc-axis. 


-l 


/    / 

\X 

f  / 

v      \ 

1/ 

\\      ° 

/    1 

/y 

For  n  =  3,  the  numbers  (8)  are  u>,  w2,  1,  which  are  repre- 
sented in  Fig.  2  by  the  vertices  of  an  equilateral  triangle. 

For  w  =  4,  R  =  cos  ir/2 +i  sin  tt/2  =  i.  The  four  fourth  roots 
of  unity  (8)  are  i,  i2=— 1,  i3=—  i,  il  =  l,  which  are  repre- 
sented by  the  vertices  of  a  square  inscribed  in  a  circle  of 
radius  unity  and  center  at  the  origin  0  (Fig,  5), 


Fig.  5 


§  11]  PRIMITIVE  ROOTS  OF  UNITY 


EXERCISES 


1.  Simplify  the  trigonometric  forms  (6)  of  the  four  fourth  roots  of  unity.  Check 
the  result  by  factoring  x4  —  1 . 

2.  For  n  =  6,  show  that  R=  —  «2.  The  sixth  roots  of  unity  are  the  three  cube  roots 
of  unity  and  their  negatives.     Check  by  factoring  x6  —  1. 

3.  From  the  point  representing  a+bi,  how  do  you  obtain  that  representing  —  (a+bi)? 
Hence  derive  from  Fig.  2  and  Ex.  2  the  points  representing  the  six  sixth  roots  of  unity. 
Obtain  this  result  another  way. 

4.  Find  the  five  fifth  roots  of  —1. 

5.  Obtain  the  trigonometric  forms  of  the  nine  ninth  roots  of  unity.  Which  of 
them  are  cube  roots  of  unity? 

6.  Which  powers  of  a  ninth  root  (7)  of  unity  are  cube  roots  of  unity? 

11.  Primitive  nth  Roots  of  Unity.  An  nth  root  of  unity  is  called 
primitive  if  n  is  the  smallest  positive  integral  exponent  of  a  power  of  it 
that  is  equal  to  unity.  Thus  p  is  a  primitive  nth  root  of  unity  if  and  only 
if  pn=l  and  pl^l  for  all  positive  integers  l<n. 

Since  only  the  last  one  of  the  numbers  (8)  is  equal  to  unity,  the  number 
R,  defined  by  (7),  is  a  primitive  nth  root  of  unity.  We  have  shown  that 
the  powers  (8)  of  R  give  all  of  the  nth  roots  of  unity.  Which  of  these 
powers  of  R  are  primitive  nth  roots  of  unity? 

For  ri  =  4,  the  powers  (8)  of  R=i  were  seen  to  be 

il=i,  i2=  —1,  i3=  —i,  ii  =  l. 

The  first  and  third  are  primitive  fourth  roots  of  unity,  and  their  exponents  1  and  3 
are  relatively  prime  to  4,  i.e.,  each  has  no  divisor  >1  in  common  with  4.  But  the 
second  and  fourth  are  not  primitive  fourth  roots  of  unity  (since  the  square  of  —1  and  the 
first  power  of  1  are  equal  to  unity),  and  their  exponents  2  and  4  have  the  divisor  2  in 
common  with  n  =  4.     These  facts  illustrate  and  prove  the  next  theorem  for  the  case 

71=4. 

Theorem.  The  primitive  nth  roots  of  unity  are  those  of  the  numbers 
(8)  whose  exponents  are  relatively  prime  to  n. 

Proof.  If  k  and  n  have  a  common  divisor  d  (d>  1),  R*  is  not  a  primitive 
nth  root  of  unity,  since 

(R*fi=(Rny=i, 

and  the  exponent  n/d  is  a  positive  integer  less  than  n. 


10  COMPLEX  NUMBERS  [Ch.  I 

But  if  k  and  n  are  relatively  prime,  i.e.,  have  no  common  divisor >1, 
Rk  is  a  primitive  nth.  root  of  unity.  To  prove  this,  we  must  show  that 
(J?ft)Ml  if  Z  is  a  positive  integer <n.     By  De  Moivre's  theorem, 

„„  2kfa  ,  .   .    2kh- 

REl  =  cos \-i  sin . 

n  n . 

If  this  were  equal  to  unity,  2klir/n  would  be  a  multiple  of  2x,  and  hence 
kl  a  multiple  of  n.  Since  k  is  relatively  prime  to  n,  the  second  factor  I 
would  be  a  multiple  of  n,  whereas  0  <  I  <  n. 

EXERCISES 

1.  Show  that  the  primitive  cube  roots  of  unity  are  w  and  «2. 

2.  For  R  given  by  (7),  prove  that  the  primitive  nth  roots  of  unity  are  (i)  for  n  =  6, 
R,  Rs;   (ii)  for  n  =  8,  R,  R3,  R5,  R7;   (Hi)  for  n  =  12,  R,  #5,  R7,  R11. 

3.  When  n  is  a  prime,  prove  that  any  nth  root  of  unity,  other  than  1,  is  primitive. 

4.  Let  R  be  a  primitive  nth  root  (7)  of  unity,  where  n  is  a  product  of  two  different 
primes  p  and  q.  Show  that  R,  .  .  .  ,  Rn  are  primitive  with  the  exception  of  Rp,  R2P,  .  .  ., 
Rap,  whose  qth.  powers  are  unity,  and  RQ,  R2a,  .  .  .  ,  Rpq,  whose  pth  powers  are  unity. 
These  two  sets  of  exceptions  have  only  Rpa  in  common.  Hence  there  are  exactly 
V1~ V~ 2+1  primitive  nth  roots  of  unity. 

5.  Find  the  number  of  primitive  nth  roots  of  unity  if  n  is  a  square  of  a  prime  p. 

6.  Extend  Ex.  4  to  the  case  in  which  n  is  a  product  of  three  distinct  primes. 

7.  If  R  is  a  primitive  15th  root  (7)  of  unity,  verify  that  R3,  R6,  R9,  R1-  are  the  primi- 
tive fifth  roots  of  unity,  and  R5  and  R10  are  the  primitive  cube  roots  of  unity.  Show 
that  their  eight  products  by  pairs  give  all  the  primitive  15th  roots  of  unity. 

8.  If  p  is  any  primitive  nth  root  of  unity,  prove  that  p,  p2,  .  .  .  ,  pn  are  distinct  and 
give  all  the  nth  roots  of  unity.  Of  these  show  that  p  is  a  primitive  nth  root  of  unity 
if  and  only  if  fc  is  relatively  prime  to  n. 

9.  Show  that  the  six  primitive  18th  roots  of  unity  are  the  negatives  of  the  primi- 
tive ninth  roots  of  unity. 


CHAPTER    II 

Elementary  Theorems  on  the  Roots  of  an  Equation 

12.  Quadratic  Equation.     If  a,  b,  c  are  given  numbers,  a^O, 

(1)  ax2+bx+c  =  0     (a  ?*0) 

is  called  a  quadratic  equation  or  equation  of  the  second  degree.  The 
reader  is  familiar  with  the  following  method  of  solution  by  "  completing 
the  square."  Multiply  the  terms  of  the  equation  by  4a,  and  transpose 
the  constant  term;  then 

4a2  x2 + 4abx  =  —  4ac. 

Adding  b2  to  complete  the  square,  we  get 

(2ax+b)2  =  A,  A  =  b2-4ac, 

,_,  -6+VA  -6-VA 

(2)  Xl=-^a—  X2  =  -^a— 

By  addition  and  multiplication,  we  find  that 

(3)  xi+x2  = ,  XiX2  =  ~. 

a  a 

Hence  for  all  values  of  the  variable  x, 

(4)  a(x— x\)  (x  —  X2)=ax2  —  a(xi-\-X2)x-)raxiX2  =  ax2+bx-\-c, 

the  sign  =  being  used  instead  of  =  since  these  functions  of  x  are  identically 
equal,  i.e.,  the  coefficients  of  like  powers  of  x  are  the  same.  We  speak 
of  a(x  —  x\)  (X—X2)  as  the  factored  form  of  the  quadratic  function  ax2-\-bx-{-c1 
and  of  x  —  X\  and  x— X2  as  its  linear  factors. 

In  (4)  we  assign  to  x  the  values  x\  and  X2  in  turn,  and  see  that 

0  =  axi2+bxi+c,     0  =  ax22-\-bx2+c. 

Hence  the  values  (2)  are  actually  the  roots  of  equation  (1). 

We  call  A  =  62  —  4ac  the  discriminant  of  the  function  ax2-\-bx-\-c  or 
of  the  corresponding  equation  (1).  If  A  =  0,  the  roots  (2)  are  evidently 
equal,  so  that,  by  (4),  ax2-\-bx-\-c  is  the  square  of  y/a(x  —  xi),  and  con- 

11 


12  THEOREMS  ON  ROOTS  OF  EQUATIONS  [Ch.  II 

versely.  We  thus  obtain  the  useful  result  that  ax2-{-bx-\-c  is  a  perfect 
square  (of  a  linear  function  of  x)  if  and  only  if  b2  =  4ac  (i.e.,  if  its  discrimi- 
nant is  zero). 

Consider  a  real  quadratic  equation,  i.e.,  one  whose  coefficients  a,  b,  c 
are  all  real  numbers.  Then  if  A  is  positive,  the  two  roots  (2)  are  real. 
But  if  A  is  negative,  the  roots  are  conjugate  imaginaries  (§2). 

When  the  coefficients  of  a  quadratic  equation  (1)  are  any  complex 
numbers,  A  has  two  complex  square  roots  (§9),  so  that  the  roots  (2)  of 
(1)  are  complex  numbers,  which  need  not  be  conjugate. 

For  example,  the  discriminant  of  x2  — 2x-\-c  is  A  =  4(l— c).  If  c  =  l,  then  A  =  0  and 
x2—2x+l  =  {x  —  V)2  is  a  perfect  square,  and  the  roots  1,  1  of  x2— 2z+l=0  are  equal. 
If  c  =  0,  A  =  4  is  positive  and  the  roots  0  and  2  of  x2— 2x=x(x  — 2)  =0  are  real.  If  c  =  2, 
A  =—4  is  negative  and  the  roots  1± V  —  1  of  x2— 2x+2  =  0  are  conjugate  complex 
numbers.     The  roots  of  x2—  x+l-\-i  =  0  are  i  and  1—  i,  and  are  not  conjugate. 

13.  Integral  Rational  Function,  Polynomial.  If  n  is  a  positive  integer 
and  Co,  ci,  .  .  .  ,  Cn  are  constants  (real  or  imaginary), 

f(x)=coxn-\-c\xn~l+\  .  .  +cn-ix+cn 

is  called  a  polynomial  in  x  of  degree  n,  or  also  an  integral  rational  function 
of  x  of  degree  n.  It  is  given  the  abbreviated  notation  f(x),  just  as  the 
logarithm  of  re +2  is  written  log  (re +2). 

If  Co  5^0,  /(re)  =0  is  an  equation  of  degree  n.  If  n  =  3,  it  is  often  called 
a  cubic  equation;  and,  if  n  =  4,  a  quartic  equation.  For  brevity,  we  often 
speak  of  an  equation  all  of  whose  coefficients  are  real  as  a  real  equation. 

14.  The  Remainder  Theorem.  If  a  polynomial  /(re)  be  divided  by 
x  —  c  until  a  remainder  independent  of  x  is  obtained,  this  remainder  is  equal 
to  /(c),  which  is  the  value  of  /(re)  when  x  =  c. 

Denote  the  remainder  by  r  and  the  quotient  by  q{x).  Since  the 
dividend  is /(re)  and  the  divisor  is  re  — c,  we  have 

f(x)  =  {x-c)  q(x)+r, 

identically  in  re.     Taking  re  =  c,  we  obtain  /(c)  =  r. 

If  r  =  0,  the  division  is  exact.  Hence  we  have  proved  also  the  follow- 
ing useful  theorem. 

The  Factor  Theorem.  If  /(c)  is  zero,  the  polynomial  /(re)  has  the 
factor  x  —  c.     In  other  words,  if  c  is  a  root  of  f(x)  =  0,  x  —  c  is  a  factor  of 


§  14]  REMAINDER  AND  FACTOR  THEOREMS  13 

For  example,  2  is  a  root  of  a;3  —  8  =  0,  so  that  x— 2  is  a  factor  of  a;3— 8.  Another 
illustration  is  furnished  by  formula  (4) 

EXERCISES 

Without  actual  division  find  the  remainder  when 
n.  x4— 3x2— x— 6  is  divided  by  x+3. 

2.  x3— 3a;2 + 6a;— 5  is  divided  by  x— 3. 
Without  actual  division  show  that 

3.  18a;10  +  19x5  +  l  is  divisible  by  x  +  1. 

4.  2x4— x3  —  6x2+4x— 8  is  divisible  by  a;— 2  and  x+2. 

5.  x4— 3a;3+3a;2  — 3x+2  is  divisible  by  x  —  1  and  x— 2. 
'6.  r3  — 1,  r4  — 1,  r5  — 1  are  divisible  by  r  —  1. 

7.  By  performing  the  indicated  multiplication,  verify  that 

rn-l  =  (r-l)  (rn~1+rn-2+  .  .  .  +r+l). 

8.  In  the  last  identity  replace  r  by  x/y,  multiply  by  yn,  and  derive 

xn-yn=(x-y)  (xw-1+xw-22/+  .  .  .  +xyn~2+yn~l). 

9.  In  the  identity  of  Exercise  8  replace  y  by  —  y,  and  derive 

xn+yn  =  (x+y)  (xn-1-xn-2y+  .  .  .  -xy^^y^1),     nodd; 
xn-yn  =  (x+y)  (xn-1-xn-2y+  .  .  .  +xyn~2 -yn~l),     neven. 

Verify  by  the  Factor  Theorem  that  x-\-y  is  a  factor. 

10.  If  o,  ar,  ar2,  .  .  .,  arn~l  are  n  numbers  in  geometrical  progression  (the  ratio  of  any 
term  to  the  preceding  being  a  constant  r^l),  prove  by  Exercise  7  that  their  sum  is 
equal  to 

a{rn-\) 
r-1     ' 

11.  At  the  end  of  each  of  n  years  a  man  deposits  in  a  savings  bank  a  dollars.  With 
annual  compound  interest  at  4%,  show  that  his  account  at  the  end  of  n  years  will  be 

—  f(1.04)re-l} 
.04 l  '         s 

dollars.  Hint:  The  final  deposit  draws  no  interest;  the  prior  deposit  will  amount  to 
a(1.04)  dollars;  the  deposit  preceding  that  will  amount  to  a(1.04)2  dollars,  etc.  Hence 
apply  Exercise  10  for  r  =  1.04. 

15.  Synthetic  Division.  The  labor  of  computing  the  value  of  a  poly- 
nomial in  x  for  an  assigned  value  of  x  may  be  shortened  by  a  simple  device. 
To  find  the  value  of 

x4+3x3-2x-5 


14  THEOREMS  ON  ROOTS  OF  EQUATIONS  [Ch.  II 

for  x  =  2,  note  that  x*=x-x3  =  2x3,  so  that  the  sum  of  the  first  two  terms 
of  the  polynomial  is  5x3.  To  5x3  =  5-22x  we  add  the  next  term  —  2x  and 
obtain  18x  or  36.  Combining  36  with  the  final  term  —5,  we  obtain  the 
desired  value  31. 

This  computation  may  be  arranged  systematically  as  follows.  After 
supplying  zero  coefficients  of  missing  powers  of  x,  we  write  the  coefficients 
in  a  line,  ignoring  the  powers  of  x. 

13  0         -2         -5    |  2 

2        10  20  36 


1        5        10  18  31 

First  we  bring  down  the  first  coefficient  1.  Then  we  multiply  it  by  the 
given  value  2  and  enter  the  product  2  directly  under  the  second  coefficient 
3,  add  and  write  the  sum  5  below.  Similarly,  we  enter  the  product  of 
5  by  2  under  the  third  coefficient  0,  add  and  write  the  sum  10  below;  etc. 
The  final  number  31  in  the  third  line  is  the  value  of  the  polynomial  when 
x  =  2.  The  remaining  numbers  in  this  third  line  are  the  coefficients,  in 
their  proper  order,  of  the  quotient 

z3+5z2+10z+18, 

which  would  be  obtained  by  the  ordinary  long  division  of  the  given  poly- 
nomial by  x  —  2. 

We  shall  now  prove  that  this  process,  called  synthetic  division,  enables 
us  to  find  the  quotient  and  remainder  when  any  polynomial  f(x)  is  divided 
by  x— c.     Write 

f(x)=a0xn-\-a1xn-1+  .  .  .  +an, 

and  let  the  constant  remainder  be  r  and  the  quotient  be 

q(x)^b0xn-1+b1xn-2+  .  .  .  +&„-i. 

By  comparing  the  coefficients  of  f(x)  with  those  in 

(x-c)  q(x)+r  =  b0xn+(b1-cbo)xn-1 

+  (b2-cbi)xn-2+  .  .  .  +(&„_!— c&„_2)a;+r— cbn-\, 

we  obtain  relations  which  become,  after  transposition  of  terms, 

&o  =  ao,    b\=ai-\-cb0,   &2  =  a2  +  c&i,  .  .  .  ,  &„_i=aB_i+c&B_2,    r  =  an-\-cb„-i. 


§  15]  SYNTHETIC  DIVISION  15 

The  steps  in  the  work  of  computing  the  &'s  may  be  tabulated  as  follows: 

Of)  0>1  0>2  ...  Cbn-l  o,n        1    c 

cbo       cbi       ...         cbn-2       cb„-i 


b0        bi         62         ...        6«_i,        r 

In  the  second  space  below  ao  we  write  60  (which  is  equal  to  ao).  We 
multiply  bo  by  c  and  enter  the  product  directly  under  a\,  add  and  write 
the  sum  61  below  it.  Next  we  multiply  61  by  c  and  enter  the  product 
directly  under  a2,  add  and  write  the  sum  62  below  it;  etc. 

EXERCISES 

Work  each  of  the  following  exercises  by  synthetic  division. 

1.  Divide  x3+3x2-2x-5  by  x-2. 

2.  Divide  2x5-x3+2x-l  by  x+2. 

3.  Divide  x3+6x2  +  10x-l  by  x-0.09. 

4.  Find  the  quotient  of  x3  —  5x2  — 2x+24  by  x— 4,  and  then  divide  the  quotient  by 
x  -  3 .     What  are  the  roots  of  x3  -  5x2  -  2x +24  =  0? 

5.  Given  that  x4— 2x3— 7x2+8x+12  =  0  has  the  roots  —1  and  2,  find  the  quadratic 
equation  whose  roots  are  the  remaining  two  roots  of  the  given  equation,  and  find  these 
roots. 

6.  If  x4  —  2x3  —  12x2  +  10x+3  =  0  has  the  roots  1  and  —3,  find  the  remaining  two  roots. 

7.  Find  the  quotient  of  2x4-x3-6x2+4x-8  by  x2-4. 

8.  Find  the  quotient  of  x4-3x3+3x2-3x+2  by  x2-3x+2. 

9.  Solve  Exercises  1,  2,  3,  6,  7  of  §  14  by  synthetic  division. 

16.  Factored  Form  of  a  Polynomial.     Consider  a  polynominal 

/(z)=c0zn+cizB-1+  .  .  .  +c  (Co 5*0), 

whose  leading  coefficient  c0  is  not  zero.  If  f(x)  =0  has  the  root  a\,  which 
may  be  any  complex  number,  the  Factor  Theorem  shows  that  f(x)  has  the 
factor  x—ai,  so  that 

f(x)  =  (x-ai)Q(x),    Q(x)=w?-1+ci'a?-a+  .  .  .  +c'a_i. 

If  Q(x)  =0  has  the  root  0:2,  then 

Q(x)  =  (x-a2)Qi(x),     f(x)  =  (x  —  ai)  (x-a2)Qi(x). 

If  Qi(x)  =0  has  the  root  0:3,  etc.,  we  finally  get 

(5)  f(x)=C0(x—ai)  (x  —  a2)  .  .  .  (x-an). 

We  shall  deduce  several  important  conclusions  from  the  preceding 
discussion.     First,  suppose  that  the  equation  fix)  =  0  of  degree  n  is  known 


16  THEOREMS  ON  ROOTS  OF  EQUATIONS  [Ch.  II 

to  have  n  distinct  roots  a\,  .  .  .  ,  an.  In  f(x)  =  (x— ai)Q(x)  take  x =02; 
then  0=(ct2— ai)  Q(«2),  whence  $(0:2)  =0  and  $(#)=()  has  the  root  0:2. 
Similarly,  Qi(rc)  =  0  has  the  root  «3,  etc.  Thus  all  of  the  assumptions 
(each  introduced  by  an  "  if  ")  made  in  the  above  discussion  have  been 
justified  and  we  have  the  conclusion  (5).  Hence  if  an.  equation  f(x)  =  0 
of  degree  n  has  n  distinct  roots  a\,  .  .  .  ,  an,  f(x)  can  be  expressed  in  the 
factored  form  (5). 

It  follows  readily  that  the  equation  can  not  have  a  root  a  different 
from  at,  .  .  .  ,  an.  For,  if  it  did,  the  left  member  of  (5)  is  zero  when 
x=a  and  hence  one  of  the  factors  of  the  right  member  must  then  be  zero, 
say  a— ctj  =  0,  whence  the  root  a  is  equal  to  aj.  We  have  now  proved 
the  following  important  result. 

Theorem.    An  equation  of  degree  n  cannot  have  more  than  n  distinct  roots. 

17.  Multiple  Roots.1  Equalities  may  occur  among  the  a's  in  (5). 
Suppose  that  exactly  mi  of  the  a's  (including  ai)  are  equal  to  a\)  that 
<x2t^oli,  while  exactly  m2  of  the  a's  are  equal  to  a.2',  etc.    Then  (5)  becomes 

(6)         f(x)  =  c0(x-ai)mi(x-a2)m2 .  .  .  (x-ak)mic,      Wi+m2+  .  .  .  +mt  =  n, 

where  ai,  .  .  .  ,  at  are  distinct.  We  then  call  a\  a  root  of  multiplicity  m\ 
of  f(x)  =  0,  0:2  a  root  of  multiplicity  W2,  etc.  In  other  words,  ai  is  a  root 
of  multiplicity  m\  of  f(x)  =  0  if  f(x)  is  exactly  divisible  by  (x— a\)mi,  but  is 
not  divisible  by  (x— ai)mi+1.  We  call  a\  also  an  mi— fold  root.  In  the 
particular  cases  wi  =  1,  2,  and  3,  we  also  speak  of  ai  as  a  simple  root,  double 
root,  and  triple  root,  respectively.  For  example,  4  is  a  simple  root,  3  a 
double  root,  —2a  triple  root,  and  6  a  root  of  multiplicity  4  (or  a  4-fold 
root)  of  the  equation 

7(3-4)  Or-3)2Or+2)3(£-6)4  =  0 

of  degree  10  which  has  no  further  root.  This  example  illustrates  the 
next  theorem,  which  follows  from  (6)  exactly  as  the  theorem  in  §  16 
followed  from  (5). 

Theorem.  An  equation  of  degree  n  cannot  have  more  than  n  roots, 
a  root  of  multiplicity  m  being  counted  as  m  roots. 

18.  Identical  Polynomials.      //  two  polynomials  in  x, 

a0xn+a1xn-1+  .  .  .  +an,         b0xn+bixn-1^-  .  .  .  +bn, 
1  Multiple  roots  are  treated  by  calculus  in  §  58. 


§20]  RELATIONS  BETWEEN   ROOTS  AND   COEFFICIENTS  17 

each  of  degree  n,  are  equal  in  value  for  more  than  n  distinct  values  of  x,  they 
are  term  by  term  identical,  i.e.,  a0  —  b0,ai  =  bi,...,  an  =  bn. 

For,  taking  their  difference  and  writing  co  =  a0  —  b0,  .  .  .  ,  cn  =  an  —  bn, 
we  have 

C0Xn  +  CXXn-l-\-   .  .  .  +Cn  =  Q 

for  more  than  n  distinct  values  of  x.  If  c05*0,  we  would  have  a  contra- 
diction with  the  theorem  in  §  16.  Hence  c0  =  0.  If  ci^O,  we  would  have 
a  contradiction  with  the  same  theorem  with  n  replaced  by  n  —  1.  Hence 
ci=0;  etc.     Thus  a0  =  bo,  ai  =  &i,  etc. 

EXERCISES 

1.  Find  a  cubic  equation  having  the  roots  0,  1,  2. 

2.  Find  a  quartic  equation  having  the  roots  ±1,  ±2. 

3.  Find  a  quartic  equation  having  the  two  double  roots  3  and  —3. 
^4.  Find  a  quartic  equation  having  the  root  2  and  the  triple  root  1. 
^S.  What  is  the  condition  that  ax2+bx-\-c  =0  shall  have  a  double  root? 

^o.  If  a0xn+  .  .  .  +an  =  0  has  more  than  n  distinct  roots,  each  coefficient  is  zero. 
7.  Why  is  there  a  single  answer  to  each  of  Exercises  1-4,  if  the  coefficient  of  the 
highest  power  of  the  unknown  be  taken  equal  to  unity?     State  and  answer  the  cor- 
responding general  question. 

19.  The  Fundamental  Theorem  of  Algebra.  Every  algebraic  equation 
with  complex  coefficients  has  a  complex  (real  or  imaginary)  root. 

This  theorem,  which  is  proved  in  the  Appendix,  implies  that  every 
equation  of  degree  n  has  exactly  n  roots  if  a  root  of  multiplicity  m  be  counted 
as  m  roots.  In  other  words,  every  integral  rational  function  of  degree  n 
is  a  product  of  n  linear  factors.  For,  in  §  16,  equations  f(x)  =0,  Q(x)  =  0, 
Qi(x)=0,  .  .  .  each  has  a  root,  so  that  (5)  and  (6)  hold. 

20.  Relations  between  the  Roots  and  the  Coefficients.     In  §  12  we 

found  the  sum  and  the  product  of  the  two  roots  of  any  quadratic  equation 
and  then  deduced  the  factored  form  of  the  equation.  We  now  apply 
the  reverse  process  to  any  equation 

(7)  f(x)=c0xn+c1xa-1+  .  .  .  +cn  =  0         (c0 5*0), 
whose  factored  form  is 

(8)  f(x)=Co(x-ai)  (x-a2)  •  •  •  (x-an). 

Our  next  step  is  to  find  the  expanded  form  of  this  product.  The  following 
special  products  may  be  found  by  actual  multiplication: 


18  THEOREMS  ON  ROOTS  OF  EQUATIONS  [Ch.  II 

(x  —  a\)  (x  —  ao)=x2—(ai-\-a2)x-\-aia2, 
(xi  —ai)  (x—a2)  (x — 03)  =  x3  —  (01  +02  -\-as)x2  +  (0102  +0:10:3 +0203)3;— 010203. 

These  identities  are  the  cases  n  =  2  and  n  =  3  of  the  following  general 

formula : 

(9)  (xi-ai)  (x-a2)  .  .  .  {x-an)=xn-{ai+  .  .  .  +on)xra_1 

+  (0102+0103+0203+  .  .  .  +ow_iora)xn_2 
,    —(010203+010204+  .  .  .  -\-an-2an-ian)xn~3 
+  .  .  .  +(— l)"oi02  .  .  .  an, 

the  quantities  in  parentheses  being  described  in  the  theorem  below.  If 
we  multiply  each  member  of  (9)  by  x— an+i,  it  is  not  much  trouble  to  verify 
that  the  resulting  identity  can  be  derived  from  (9)  by  changing  n  into 
n+1,  so  that  (9)  is  proved  true  by  mathematical  induction.  Hence  the 
quotient  of  (7)  by  c0  is  term  by  term  identical  with  (9),  so  that 

01+02+  .  .  .  +o„=—  ci/c0, 

O1O2+O103+O203+   .   .  .   +Ora_iO„  =  C2/Co, 

(10) 

010203+010204+    .    .   .    +0„_20„_lOn=  —  C3/C0, 

0102  .  .  .  o„_ia:n=(  —  l)wC„/c0. 
These  results  may  be  expressed  in  the  following  words: 

Theorem.  If  a\,  .  .  .  ,  an  are  the  roots  of  equation  (7),  the  sum  of  the 
roots  is  equal  to  —  ci/c0,  the  sum  of  the  products  of  the  roots  taken  two  at 
a  time  is  equal  to  C2/C0,  the  sum  of  the  products  of  the  roots  taken  three  at  a 
time  is  equal  to  —cs/cq,  etc.;  finally,  the  product  of  all  the  roots  is  equal  to 
(-l)ncn/c0. 

Since  we  may  divide  the  terms  of  our  equation  (7)  by  Cq,  the  essential 
part  of  our  theorem  is  contained  in  the  following  simpler  statement: 

Corollary.  In  an  equation  in  x  of  degree  n,  in  which  the  coefficient 
of  xn  is  unity,  the  sum  of  the  n  roots  is  equal  to  the  negative  of  the  coefficient 
of  xn~l,  the  sum  of  the  products  of  the  roots  two  at  a  time  is  equal  to  the  coeffi- 
cient of  xn~2,  etc.;  finally  the  product  of  all  the  roots  is  equal  to  the  constant 
term  or  its  negative,  according  as  n  is  even  or  odd. 

For  example,  in  a  cubic  equation  having  the  roots  2,  2,  5,  and  having  unity  as  the 
coefficient  of  xz,  the  coefficient  of  x  is  2  •  2  +2  •  5  +2  •  5  =  24. 


§  21]  IMAGINARY  ROOTS  OCCUR  IN  PAIRS  19 

EXERCISES 

1.  Find  a  cubic  equation  having  the  roots  1,  2,  3. 

2.  Find  a  quartic  equation  having  the  double  roots  2  and  —2. 

3.  Solve  x4— 6x3+13x2  — 12x+4  =  0,  which  has  two  double  roots. 

4.  Prove  that  one  root  of  x3+px2+gx+r  =  0  is  the  negative  of  another  root  if  and 
only  if  r  =  pq. 

i  5.  Solve  4x3  — 16x2  — 9x+36  =  0,  given  that  one  root  is  the  negative  of  another. 

6.  Solve  x3—  9x2+23x  — 15=0,  given  that  one  root  is  the  triple  of  another. 

7.  Solve  x4— 6x3+12x2  —  10x+3  =  0,  which  has  a  triple  root. 

8.  Solve  x3  —  14x2— 84x+216  =  0,  whose  roots  are  in  geometrical  progression,  i.e., 
with  a  common  ratio  r  [say  m/r,  m,  mr]. 

9.  Solve  x3  —  3x2  —  13x+15  =  0,  whose  roots  are  in  arithmetical  progression,  i.e., 
with  a  common  difference  d  [say  m  —  d,  m,  rn+d]. 

10.  Solve  x4— 2x3  — 21x2+22x+40  =  0,  whose  roots  are  in  arithmetical  progression. 
[Denote  them  by  c  — 36,  c  —  b,  c+b,  c+36,  with  the  common  difference  2&  ] 

11.  Find  a  quadratic  equation  whose  roots  are  the  squares  of  the  roots  of 

x2  —  px+g  =  0. 

12.  Find  a  quadratic  equation  whose  roots  are  the  cubes  of  the  roots  of  x2  —  px+q  =  0. 
Hint:  a3+j33  =  (a+/3)3-3a/3(a+/3). 

13 .  If  a  and  /3  are  the  roots  of  x*  —  px +q  =  0,  find  an  equation  whose  roots  are  (i)  a2//3 ; 
and  p2/a;   (ii)  a3/3  and  a/33;   (iii)  a+l//3  and  /3  +  1/a. 

^14.  Find  a  necessary  and  sufficient  condition  that  the  roots,  taken  in  some  order, 
of  x3+px2+gx+r  =  0  shall  be  in  geometrical  progression. 
15'.  Solve  x3-28x+48  =  0,  given  that  two  roots  differ  by  2. 

21.  Imaginary  Roots  occur  in  Pairs.  The  two  roots  of  a  real  quadratic 
equation  whose  discriminant  is  negative  are  conjugate  imaginaries  (§  12). 
This  fact  illustrates  the  following  useful  result. 

Theorem.  If  an  algebraic  equation  with  real  coefficients  has  the  root  a-\-bi, 
where  a  and  b  are  real  and  b^O,  it  has  also  the  root  a  —  bi. 

Let  the  equation  be  f(x)  =  0  and  divide  f(x)  by 

(11)  (x-a)2+b2=(x-a-bi)  {x-a+bi) 

until  we  reach  a  remainder  rx-\-s  whose  degree  in  x  is  less  than  the  degree 
of  the  divisor.  Since  the  coefficients  of  the  dividend  and  divisor  are  all 
real,  those  of  the  quotient  Q(x)  and  remainder  are  real.     We  have 

f(x)=Q{x)  { (x-af+b2}  +rx+s, 

identically  in  x.  This  identity  is  true  in  particular  when  x  =  a-\-bi,  so 
that 

0  =  r(a  -\-bi)  -\-s  =  ra-\-s-\-rbi. 


20  THEOREMS  ON  ROOTS  OF  EQUATIONS  [Ch.  II 

Since  all  of  the  letters,  other  than  i,  denote  real  numbers,  we  have  (§2) 
ra+s  =  0,  rb  =  0.  But  6^0.  Hence  r  =  0,  and  then  s  =  0.  Hence  f(x) 
is  exactly  divisible  by  the  function  (11),  so  that  f(x)  =  0  has  the  root  a—bi. 

The  theorem  may  be  applied  to  the  real  quotient  Q(x).  We  obtain 
the 

Corollary.  If  a  real  algebraic  equation  has  an  imaginary  root  of 
multiplicity  m,  the  conjugate  imaginary  of  this  root  is  a  root  of  multiplicity  m. 

Counting  a  root  of  multiplicity  masm roots,  we  see  that  a  real  equation 
cannot  have  an  odd  number  of  imaginary  roots.  Hence  by  §19,  a  real 
equation  of  odd  degree  has  at  least  one  real  root. 

Of  the  n  linear  factors  of  a  real  integral  rational  function  of  degree  n 
(§  19),  those  having  imaginary  coefficients  may  be  paired  as  in  (11). 
Hence  every  integral  rational  function  with  real  coefficients  can  be  expressed 
as  a  product  of  real  linear  and  real  quadratic  factors. 

EXERCISES 

1.  Solve  x3- 3x2 -6x- 20  =  0,  one  root  being  -l+V-3. 
v2.  Solve  x4  —  4x3-\-5x2  —  2x— 2  =  0,  one  root  being  1—  i. 

3.  Find  a  cubic  equation  with  real  coefficients  two  of  whose  roots  are  1  and  3+2i. 

4.  If  a  real  cubic  equation  x3—  6x2+  ...  =0  has  the  root  1  +  V  —  5,  what  are  the 
semaining  roots?     Find  the  complete  equation. 

'h.  If  an  equation  with  rational  coefficients  has  a  root  o+  v  b,  where  a  and  b  are 
rational,  but  Vfo  is  irrational,  prove  that  it  has  the  root  a—vb.  [Use  the  method  of 
§21.] 

6.  Solve  x4— 4x3+4x  — 1  =0,  one  root  being  2  +  V  3. 

7.  Solve  x3-(4+V/3)x2  +  (5+4V/3)x-5V/3=0,  having  the  root  Vi". 

8.  Solve  the  equation  in  Ex.  7,  given  that  it  has  the  root  2+i. 

9.  Find  a  cubic  equation  with  rational  coefficients  having  the  roots  §,  |+v2. 

10.  Given  that  x4  —  2x3  —  5x2—&x-{-2  —  0  has  the  root  2  —  V  3,  find  another  root  and 
by  means  of  the  sum  and  the  product  of  the  four  roots  deduce,  without  division,  the 
quadratic  equation  satisfied  by  the  remaining  two  roots. 

11.  Granted  that  a  certain  cubic  equation  has  the  root  2  and  no  real  root  different 
from  2,  does  it  have  two  imaginary  roots? 

12.  Granted  that  a  certain  quartic  equation  has  the  roots  2±3t,  and  no  imaginary 
roots  different  from  them,  does  it  have  two  real  roots? 

13.  By  means  of  the  proof  of  Ex.  5,  may  we  conclude  as  at  the  end  of  §  21  that 
every  integral  rational  function  with  rational  coefficients  can  be  expressed  as  a  product 
of  linear  and  quadratic  factors  with  rational  coefficients? 


§  22]  UPPER  LIMIT  TO  REAL  ROOTS  21 

22.  Upper  Limit  to  the  Real  Roots.  Any  number  which  exceeds 
all  real  roots  of  a  real  equation  is  called  an  upper  limit  to  the  real  roots. 
We  shall  prove  two  theorems  which  enable  us  to  find  readily  upper  limits 
to  the  real  roots.  For  some  equations  Theorem  I  gives  a  better  (smaller) 
upper  limit  than  Theorem  II;  for  other  equations,  the  reverse  is  true. 
Evidently  any  positive  number  is  an  upper  limit  to  the  real  roots  of  an 
equation  having  no  negative  coefficients. 

Theorem  I.     If,  in  a  real  equation 

f(x)=a0xn+aiXn-1-\-  .  .  .  +a„  =  0     (a0>0), 

the  first  negative  coefficient  is  preceded  by  k  coefficients  which  are  positive  or 
zero,  and  if  G  denotes  the  greatest  of  the  numerical  values  of  the  negative 
coefficients,  then  each  real  root  is  less  than  l  +  -y/(r/ao. 

For  example,  in  x5+4x4  —  7x2  —  40a; +1=0,  G  =  40  and  k  =  3  since  we  must  supply 
the  coefficient  zero  to  the  missing  power  z3.  Thus  the  theorem  asserts  that  each  root 
is  less  than  1  +  V40  and  therefore  less  than  4.42.  Hence  4.42  is  an  upper  limit  to  the 
roots. 

Proof.  For  positive  values  of  x,  f{x)  will  be  reduced  in  value  or  remain 
unchanged  if  we  omit  the  terms  a\xn~l,  .  .  .  ,  ak-ixn~k+1  (which  are 
positive  or  zero),  and  if  we  change  each  later  coefficient  ak,  .  .  .  ,  an  to 
— G.     Hence 

f(x)^a0xn-G(xn-k-\-xn-k-1-{-  .  .  .  +x+l). 

But,  by  Ex.  7  of  §  14, 

zw-*+  •  ••  +X+l=- — -, 

x—  1 

ifz^l.     Furthermore, 

a°x~G\     x-i     )=-  -^=r~ 

Hence,  if  x>l, 

xn-k+1{a0xk-\x-l)-G} 


f{x)> 


x—l 

xn-k+1{a0(x-l)k-G} 


x-1 

Thus,  for  x>l,  f(x)>0  and  x  is  not  a  root   if  a0  (x—l)k  —  G^.0,  which 
is  true  if  x|i  l  +  ^G'/ao. 

THE  CATHERINE  B.  O'CONNOR  LIBRARY 
!  03SL 
WESTON,  MASSACHUSETTS    02193 


22  THEOREMS  ON  ROOTS  OF  EQUATIONS  [Ch.  II 

23.  Another  Upper  Limit  to  the  Roots. 

Theokem  II.  //,  in  a  real  algebraic  equation  in  which  the  coefficient 
of  the  highest  power  of  the  unknown  is  positive,  the  numerical  value  of  each 
negative  coefficient  be  divided  by  the  sum  of  all  the  positive  coefficients  which 
precede  it,  the  greatest  quotient  so  obtained  increased  by  unity  is  an  upper 
limit  to  the  roots. 

For  the  example  in  §  22,  the  quotients  are  7/(1+4)  and  40/5,  so  that  Theorem  II 
asserts  that  1+8  or  9  is  an  upper  limit  to  the  roots.  Theorem  I  gave  the  better  upper 
limit  4.42.  But  for  x3+8x2  — 9x+c2  =  0,  Theorem  I  gives  the  upper  limit  4,  while 
Theorem  II  gives  the  better  upper  limit  2. 

We  first  give  the  proof  for  the  case  of  the  equation 

/(x)  =p4xi—p3x3+p2X2—p1x+po  =  0 

in  which  each  pt  is  positive.     In  view  of  the  identities 

x4  =  (x-l)  (x3+x2+x+l)+l,        x2  =  (x-l)  (x  +  l)+l, 

/  (x)  is  equal  to  the  sum  of  the  terms 

Pi(x  —  l)x3  +pi(x  —  l)x2+pi(x  —  l)x +pi(x  —  1)  +p4, 

— Pzxz  +p2(x  —  l)x+p2(x  —  1)+P2, 

—pix  +p0- 

If  x>l,  negative  terms  occur  only  in  the  first  and  third  columns,  while  the  sum  of  the 
terms  in  each  of  these  two  columns  will  be  ^L  0  if 

P4(x-1)-P8^0,    (p4+p2)  (z-D-pi^O. 

Hence  f(x)  >  0  and  x  is  not  a  root  if 

»>!+&    *>1+-  Pl 


Pi  Pi  +  P2 

This  proves  the  theorem  for  the  present  equation. 

Next,  let/(x)  be  modified  by  changing  its  constant  term  to  —  p0.  We  modify  the 
above  proof  by  employing  the  sum  (pi-\-pi)x  —  p0  of  all  the  terms  in  the  corresponding 
last  two  columns.     This  sum  will  be  >0  if  x>p0/(P4+P2),  which  is  true  if 

*>l+-i^ 


P4  +  P2 

To  extend  this  method  of  proof  to  the  general  case 

f(x)=anxn+  .  .  .  +a0         (an>0), 

we  have  only  to  employ  suitable  general  notations.     Let  the  negative 
coefficients  be    atl,  .  .  .  ,  att,   where    &i>/c2>  •  •  •  >fa-     For    each    posi- 


§  23]  UPPER  LIMIT  TO  REAL  ROOTS  23 

tive  integer  m  which  is  £  n  and  distinct  from  fci,  .  .  .  ,  kt}  we  replace  xm 
by  the  equal  value 

d(xm~l+xm-2+  .  .  .  +a;+l)+l 

where  d=z— 1.  Let  F(x)  denote  the  polynomial  in  x,  with  coefficients 
involving  d,  which  is  obtained  from  f(x)  by  these  replacements.  Let 
x>l,  so  that  d  is  positive.  Thus  the  terms  a^xh  are  the  only  negative 
quantities  occurring  in  F(x).  If  kt>0,  the  terms  of  F(x)  which  involve 
explicitly  the  power  xh  are  a^xh  and  the  amdxh  for  the  various  positive 
coefficients  am  which  precede  a4<.  The  sum  of  these  terms  will  be  ^.0 
if  a^+dSa^^O,  i.e.,  if 

—at. 

There  is  an  additional  case  if  kt  =  0,  i.e.,  if  a0  is  negative.  Then  the 
terms  of  F(x)  not  involving  x  explicitly  are  a0  and  the  am(d-{-l)  for  the 
various  positive  coefficients  am.      Their  sum,  ao-\-x2,am,  will  be   >0  if 

—  «o 

which  is  true  if 

EXERCISES 

Apply  the  methods  of  both  §  22  and  §  23  to  find  an  upper  limit  to  the  roots  of 

1.  4x5-8x4+22x3+98.r2-73a;+5=0. 

2.  z4-5x3+7x2-8x+l=0. 

3.  27+3x6-4x5+5x4-6x3-7x2-8=0. 

4.  x7+2x5+4x4-8x2-32=0. 

5.  A  lower  limit  to  the  negative  roots  of  /(x)=0  may  be  found  by  applying  our 
theorems  to/(—  x)  =0,  i.e.,  to  the  equation  derived  from/(x)  =0  by  replacing  x  by  — x. 
Find  a  lower  limit  to  the  negative  roots  in  Exs.  2,  3,  4. 

6.  Prove  that  every  real  root  of  a  real  equation  /(x)  =0  is  less  than  \-\-g/aQ  if  a0>0, 
where  g  denotes  the  greatest  of  the  numerical  values  of  a\,  .  .  .  ,  an.    Hint:   if  x>0, 

a0xn+a1xn-1+  .  .  .^a0xn-g(xn-1+  .  .  .  +x+l). 

Proceed  as  in  §  22  with  k  =  1 . 

7.  Prove  that  l+gr^|ac|  is  an  upper  limit  for  the  moduli  of  all  complex  roots  of  any 
equation  /(x)=0  with  complex  coefficients,  where  g  is  the  greatest  of  the  values  |oi|, 
.  .  .  ,  \an\,  and  \a\  denotes  the  modulus  of  a.     Hint:  use  Ex.  5  of  §  8. 


24  THEOREMS  ON  ROOTS  OF  EQUATIONS  [Ch.  II 

24.  Integral  Roots.     For  an  equation  all  of  whose  coefficients  are  integers, 
any  integral  root  is  an  exact  divisor  of  the  constant  term. 
For,  if  x  is  an  integer  such  that 

(12)  a0xn+  .  .  .  +an-ix+an  =  0, 

where  the  a's  are  all  integers,  then,  by  transposing  terms,  we  obtain 

x{— aQxn~l—  .  .  .  —an-i)=an. 

Thus  x  is  an  exact  divisor  of  an  since  the  quotient  is  the  integer  given  by 
the  quantity  in  parenthesis. 

Example  1.     Find  all  the  integral  roots  of 

x3+x2-3x+9=0. 

Solution.     The  exact  divisors  of  the  constant  term  9  are  ±1,  ±3,  ±9.     By  trial, 
no  one  of  ±1,  3  is  a  root.     Next,  we  find  that  —3  is  a  root  by  synthetic  division  (§  15) : 

1         1-3        9     1-3 
-3        6     -9 


1-230 

Hence  the  quotient  is  z2-2x+3,  which  is  zero  for  x  =  l±V -2.  Thus  —3  is  the 
only  integral  root. 

When  the  constant  term  has  numerous  exact  divisors,  some  device 
may  simplify  the  application  of  the  theorem. 

Example  2.1     Find  all  the  integral  roots  of 

2/3+12y2-32y -256  =  0. 

Solution.  Since  all  the  terms  except  y3  are  divisible  by  2,  an  integral  root  y  must 
be  divisible  by  2.  Since  all  the  terms  except  y3  are  now  divisible  by  24,  we  have  y  =  4z, 
where  z  is  an  integer.     Removing  the  factor  26  from  the  equation  in  z,  we  obtain 

z3+3z2-2z-4=0. 

An  integral  root  must  divide  the  constant  term  4.  Hence,  if  there  are  any  integral 
roots,  they  occur  among  the  numbers  ±1,  ±2,  ±4.  By  trial,  —1  is  found  to  be  a 
root: 

1         3-2-4     1-1 
-1-2         4 


12-40 
1  This  problem  is  needed  for  the  solution  (§  48)  of  a  certain  quartic  equation. 


§  24]  INTEGRAL  ROOTS  25 

Hence  the  quotient  is  z2+2z— 4,  which  is  zero  for  z=-l±V5.  Thus  y=4z=—  4 
is  the  only  integral  root  of  the  proposed  equation. 

EXERCISES 

Find  all  the  integral  roots  of 

1.  x3+8x2+13x+6  =  0.  2.  x3-5x2-2x+24  =  0. 

3.  x3-10x2+27x-18=0.  4.  x4+4x3+8x+32=0. 
5.  The  equation  in  Ex.  4  of  §  23. 

25.  Newton's  Method  for  Integral  Roots.  In  §  24  we  proved  that 
an  integral  root  x  of  equation  (12)  having  integral  coefficients  must  be 
an  exact  divisor  of  an.  Similarly,  if  we  transpose  all  but  the  last  two 
terms  of  (12),  we  see  that  an-\x-\-an  must  be  divisible  by  x2,  and  hence 
an-\-\-an/x  divisible  by  x.  By  transposing  all  but  the  last  three  terms 
of  (12),  we  see  that  their  sum  must  be  divisible  by  xs,  and  hence  a„_2+ 
(an-i-\-an/x)/x  divisible  by  x.  We  thus  obtain  a  series  of  conditions 
of  divisibility  which  an  integral  root  must  satisfy.  The  final  sum 
a0-{-ai/x-\-  .  .  .  must  not  merely  be  divisible  by  x,  but  be  actually  zero, 
since  it  is  the  quotient  of  the  function  (12)  by  x11. 

In  practice,  we  must  test  in  turn  the  various  divisors  x  of  an.  If  a 
chosen  x  is  not  a  root,  that  fact  will  be  disclosed  by  one  of  the  conditions 
mentioned.  Newton's  method  is  quicker  than  synthetic  division  since 
it  usually  detects  early  and  throws  out  wrong  guesses  as  to  a  root,  whereas 
in  synthetic  division  the  decision  comes  only  at  the  final  step. 

For  example,  the  divisor  —3  of  the  constant  term  of 

(13)  /(x)=x4-9x3+24x2-23x  +  15=0 

is  not  a  root  since  — 23+15/(— 3)  =  —28  is  not  divisible  by  —3.  To  show  that  none 
of  the  tests  fails  for  3,  so  that  3  is  a  root,  we  may  arrange  the  work  systematically  as 
f  ollows : 

3 


1     -9         24     -23     15 
(14)  -16-6  5 


(divisor) 


0     -3         18     -18 


First  we  divide  the  final  coefficient  15  by  3,  place  the  quotient  5  directly  under  the  coef- 
ficient —23,  and  add.  Next,  we  divide  this  sum  —18  by  3,  place  the  quotient  —6 
directly  under  the  coefficient  24,  and  add.  After  two  more  such  steps  we  obtain  the 
sum  zero,  so  that  3  is  a  root. 

It  is  instructive  to  obtain  the  preceding  process  by  suitably  modifying  synthetic 
division.     First,  we  replace  x  by  l/y  in  (13),  multiply  each  term  by  y4,  and  obtain 

15?/4-23y3+24?/2-9?/  +  l  =0. 


26  THEOREMS  ON  ROOTS  OF  EQUATIONS  [Ch.  II 

We  may  test  this  for  the  root  y  =  \,  which  corresponds  to  the  root  x  =  3  of  (13),  by 
ordinary  synthetic  division: 

15     -23        24     -9        1  i 

5       —6         6—1  (multiplier) 


15     -18         18     -3         0 


The  coefficients  in  the  last  two  lines  (after  omitting  15)  are  the  same  as  those  of  the  last 
two  fines  in  (14)  read  in  reverse  order.  This  should  be  the  case  since  we  have  here 
multiplied  the  same  numbers  by  ^  that  we  divided  by  3  in  (14).  The  numbers  in 
the  present  third  line  are  the  coefficients  of  the  quotient  (§  15).  Since  we  equate  the 
quotient  to  zero  for  the  applications,  we  may  replace  these  coefficients  by  the  numbers 
in  the  second  line  which  are  the  products  of  the  former  numbers  by  -§-.  The  numbers 
in  the  second  line  of  (14)  are  the  negatives  of  the  coefficients  of  the  quotient  of  f(x) 
by  x-Z. 

Example.     Find  all  the  integral  roots  of  equation  (13). 

Solution.  For  a  negative  value  of  x,  each  term  is  positive.  Hence  all  the  real 
roots  are  positive.  By  §  23,  10  is  an  upper  limit  to  the  roots.  By  §  24,  any  integral 
root  is  an  exact  divisor  of  the  constant  term  15.  Hence  the  integral  roots,  if  any,  occur 
among  the  numbers  1,  3,  5.  Since /(l)  =8,  1  is  not  a  root.  By  (14),  3  is  a  root.  Pro- 
ceeding similarly  with  the  quotient  by  a;— 3,  whose  coefficients  are  the  negatives  of  the 
numbers  in  the  second  line  of  (14),  we  find  that  5  is  a  root. 

EXERCISES 

1.  Solve  Exs.  1-4  of  §  24  by  Newton's  method. 

2.  Prove  that,  in  extending  the  process  (14)  to  the  general  equation  (12),  we  may 
employ  the  final  equations  in  §  15  with  r  =  0  and  write 


c 
(divisor) 


flo  Q>i  O2     •  •  •  0,n — 2  ®"n — 1      ' 

—  b0      —hi      —62     .  .  .      —bn-2      —bn-i 

0     —  cb0    —cbi     .  .  .    —  cb„—2    —cbn—2 

Here  the  quotient,  —bn—i,  of  an  by  c  is  placed  directly  under  an— 1;  and  added  to  it  to 
yield  the  sum  —cbn-2,  etc. 

26.  Another  Method  for  Integral  Roots.  An  integral  divisor  d  of  the 
constant  term  is  not  a  root  if  d—m  is  not  a  divisor  of  f(m),  where  m  is 
any  chosen  integer.     For,  if  d  is  a  root  of  f(x)  =  0,  then 

f(x)m(z-d)Q(x), 

where  Q(x)  is  a  polynomial  having  integral  coefficients  (§  15).     Hence 
f(m)  =  (m  —  d)  q,  where  q  is  the  integer  Q(ra). 


§  27]  RATIONAL  ROOTS  27 

In  the  example  of  §  25,  take  d  =  15,  m  =  1.  Since /(l)  =  8  is  not  divisible  by  15  —  1  =  14, 
15  is  not  an  integral  root. 

Consider  the  more  difficult  example 

fix)  =x3  -20x2+164z-400  =  0, 

whose  constant  term  has  many  divisors.  There  is  evidently  no  negative  root,  while 
21  is  an  upper  limit  to  the  roots.  The  positive  divisors  less  than  21  of  400  =  2452  are 
d  =  1,  2,  4,  8,  16,  5,  10,  20.  First,  take  m  =  1  and  note  that/(l)  =  -255  =  -3  •  5  •  17.  The 
corresponding  values  of  d  —  1  are  0,  1,  3,  7,  15,  4,  9,  19;  of  these,  7,  4,  9,  19  are  not  divisors 
of /(l),  so  that  d  =  S,  5,  10  and  20  are  not  roots.  Next,  take  m  =  2  and  note  that  /(2)  = 
—  144  is  not  divisible  by  16—2  =  14.  Hence  16  is  not  a  root.  Incidentally,  d  =  l  and 
d  =  2  were  excluded  since /(d)  ^0.     There  remains  only  d  =  4,  which  is  a  root. 

In  case  there  are  numerous  divisors  within  the  limits  to  the  roots,  it 
is  usually  a  waste  of  time  to  list  all  these  divisors.  For,  if  a  divisor  is 
found  to  be  a  root,  it  is  preferable  to  employ  henceforth  the  quotient, 
as  was  done  in  the  example  in  §  25. 

EXERCISES 

Find  all  the  integral  roots  of 

1.  z4-2z3-21a;2+22a;+40  =  0. 

2.  y3-9y2 -24^+216  =  0. 

3.  x"-23x3+187a;2-653x+936  =  0. 

4.  z5+47x4+423x3  +  140x2+1213x-420  =  0. 

5.  x5 -34a;3 +29x2+212x -300  =  0. 

27.  Rational  Roots.     If  an  equation  with  integral  coefficients 

(15)  c0a;M+cix"-1+  .  .  .  +c„_ia;+c„  =  0 

has  the  rational  root  a/b,  where  a  and  b  are  integers  without  a  common  divisor 
>  1,  then  a  is  an  exact  divisor  of  cn,  and  b  is  an  exact  divisor  of  Cq. 

Insert  the  value  a/b  of  x  and  multiply  all  terms  of  the  equation  by 
bn.     We  obtain 

c0an+c1an~1b+  .  .  .  -\-cn-1abn-1+cnbn  =  0. 

Since  a  divides  all  the  terms  preceding  the  last  term,  it  divides  that  term. 
But  a  has  no  divisor  in  common  with  bn;  hence  a  divides  cn.  Similarly, 
b  divides  all  the  terms  after  the  first  term  and  hence  divides  c0. 

Example.     Find  all  the  rational  roots  of 

2z3-7x2+10x-6=0. 


28  THEOREMS  ON  ROOTS  OF  EQUATIONS  [Ch.  II 

Solution.     By  the  theorem,  the  denominator  of  any  rational  root  £  is  a  divisor  of  2. 
Hence  y  =  2x  is  an  integer.     Multiplying  the  terms  of  our  equation  by  4,  we  obtain 

y3-7y2+20y-24  =  0. 

There  is  evidently  no  negative  root.  By  either  of  the  tests  in  §§  22,  23,  an  upper  limit 
to  the  positive  roots  of  our  equation  in  x  is  1+7/2,  so  that  t/<9.  Hence  the  only 
possible  values  of  an  integral  root  y  are  1,  2,  3,  4,  6,  8.  Since  1  and  2  are  not  roots,  we 
try  3: 

1     -7      20     -24     |_3_ 
-1        4     -8 


0     -3       12 


Hence  3  is  a  root  and  the  remaining  roots  satisfy  the  equation  y2—  4y+8  =  0  and  are 
2±2£.     Thus  the  only  rational  root  of  the  proposed  equation  is  x  =  3/2. 

If  c0  =  l,  then  6=  ±1  and  a/b  is  an  integer.     Hence  we  have  the 
Corollary.     Any  rational  root  of  an  equation  with  integral  coefficients, 
that  of  the  highest  power  of  the  unknown  being  unity,  is  an  integer. 
Given  any  equation  with  integral  coefficients 

aQyn-^-a1yn-1-\-  .  .  .  +aw  =  0, 

we  multiply  each  term  by  aQn~l,  write  a0y  =  x,  and  obtain  an  equation 
(15)  with  integral  coefficients,  in  which  the  coefficient  c0  of  xn  is  now  unity. 
By  the  Corollary,  each  rational  root  x  is  an  integer.  Hence  we  need  only 
find  all  the  integral  roots  x  and  divide  them  by  ao  to  obtain  all  the  rational 
roots  y  of  the  proposed  equation. 

Frequently  it  is  sufficient  (and  of  course  simpler)  to  set  ky  =  x, where 
A;  is  a  suitably  chosen  integer  less  than  a0. 

EXERCISES 

Find  all  of  the  rational  roots  of 

1-  V'-L i/2/3+J-|o^-402/+9  =  0. 

2.  6y3-ll2/2+62/-l=0. 

3.  108?/3-270?/2-422/  +  l=0.     [Use/b  =  6.] 

4.  32i/3-6f/-l=0.     [Use  the  least  k.} 

5.  96?/3-16?/2-62/+l=0.  6.  247/3-2?/2-52/  +  l  =0. 
7.  y3-±y2-2y+l=0.                              <  8.  y*-%y*+3y -2=0. 
9.  Solve  Exs.  2-0  by  replacing  y  by  1/x. 

Find  the  equations  whose  roots  are  the  products  of  0  by  the  roots  of 
10.  2/2-2y-L  =  0.  11.  2/3—b2-!2/+i  =  0. 


CHAPTER  III 


Constructions  with  Ruler  and  Compasses 

28.  Impossible  Constructions.  We  shall  prove  that  it  is  not  possible, 
by  the  methods  of  Euclidean  geometry,  to  trisect  all  angles,  or  to  con- 
struct a  regular  polygon  of  7  or  9  sides.  The  proof,  which  is  beyond  the 
scope  of  elementary  geometry,  is  based  on  principles  of  the  theory  of 
equations.  Moreover,  the  discussion  will  show  that  a  regular  polygon 
of  17  sides  can  be  constructed  with  ruler  and  compasses,  a  fact  not  suspected 
during  the  twenty  centuries  from  Euclid  to  Gauss. 


29.  Graphical  Solution  of  a  Quadratic  Equation. 

structible,  and 


If  a  and  b  are  con- 


(1)  x2-ax+b  =  0 

has  real  coefficients  and  real  roots,  the  roots 
can  be  constructed  with  ruler  and  compasses 
as  follows.  Draw  a  circle  having  as  a  diam- 
eter the  line  BQ  joining  the  points  B  = 
(0,  1)  and  Q=(a,  b)  in  Fig.  6.  Then  the 
abscissas  ON  and  OM  of  the  points  of  inter- 
section of  this  circle  with  the  x-axis  are  the 
roots  of  (1). 

For,  the  center  of  the  circle  is  (a/2,  (6+l)/2);  the  square  of  BQ  is 
a2 +(6—  l)2;  hence  the  equation  of  the  circle  is 


Fig.  6 


hh^j- 


a2+(b-lf 


2    /  4 

This  is  found  to  reduce  to  (1)  when  y  =  0,  which  proves  the  theorem. 

When  the  circle  is  tangent  to  the  x-axis,  so  that  M  and  N  coincide, 
the  two  roots  are  equal.  When  the  circle  does  not  cut  the  rc-axis,  or 
when  Q  coincides  with  B,  the  roots  are  imaginary. 

Another  construction  follows  from  §  30. 

29 


30  CONSTRUCTIONS  WITH  RULER  AND  COMPASSES  [Ch.  Ill 

EXERCISES 

Solve  graphically: 

1.  x2-5x+4  =  0.  2.  x2+5x+4  =  0.  3.  z2+5z-4  =  0. 

4.  x2-5a;-4  =  0.  5.  a;2-4x+4  =  0.  6.  x2-3x+4  =  0. 

30.  Analytic  Criterion  for  Constructibility.  The  first  step  in  our 
consideration  of  a  problem  proposed  for  construction  consists  in  formu- 
lating the  problem  analytically.  In  some  instances  elementary  algebra 
suffices  for  this  formulation.  For  example,  in  the  ancient  problem  of 
the  duplication  of  a  cube,  we  take  as  a  unit  of  length  a  side  of  the  given 
cube,  and  seek  the  length  x  of  a  side  of  another  cube  whose  volume  is 
double  that  of  the  given  cube;  hence 

(2)  x3  =  2. 

But  usually  it  is  convenient  to  employ  analytic  geometry  as  in  §  29; 
a  point  is  determined  by  its  coordinates  x  and  y  with  reference  to  fixed 
rectangular  axes;  a  straight  line  is  determined  by  an  equation  of  the 
first  degree,  a  circle  by  one  of  the  second  degree,  in  the  coordinates  of  the 
general  point  on  it.  Hence  we  are  concerned  with  certain  numbers, 
some  being  the  coordinates  of  points,  others  being  the  coefficients  of  equa- 
tions, and  still  others  expressing  lengths,  areas  or  volumes.  These  num- 
bers may  be  said  to  define  analytically  the  various  geometric  elements 
involved. 

Criterion.  A  proposed  construction  is  possible  by  ruler  and  com- 
passes if  and  only  if  the  numbers  which  define  analytically  the  desired  geo- 
metric elements  can  be  derived  from  those  defining  the  given  elements  by  a 
finite  number  of  rational  operations  and  extractions  of  real  square  roots. 

In  §  29  we  were  given  the  numbers  a  and  b,  and  constructed  lines  of  lengths 

^(a±Va2-46). 

Proof.  First,  we  grant  the  condition  stated  in  the  criterion  and  prove 
that  the  construction  is  possible  with  ruler  and  compasses.  For,  a  rational 
function  of  given  quantities  is  obtained  from  them  by  additions,  sub- 
tractions, multiplications,  and  divisions.  The  construction  of  the  sum 
or  difference  of  two  segments  is  obvious.  The  construction,  by  means 
of  parallel  fines,  of  a  segment  whose  length  p  is  equal  to  the  product  a-b 
of  the  lengths  of  two  given  segments  is  shown  in  Fig.  7;  that  for  the  quo- 


30] 


ANALYTIC  CRITERION  FOR  CONSTRUCTIBILITY 


31 


tient  q  =  a/b  in  Fig.  8.  Finally,  a  segment  of  length  s  =  Vn  may  be  con- 
structed, as  in  Fig.  9,  by  drawing  a  semicircle  on  a  diameter  composed 
of  two  segments  of  lengths  1  and  n,  and  then  drawing  a  perpendicular 
to  the  diameter  at  the  point  which  separates  the  two  segments.  Or  we 
may  construct  a  root  of  x2  —  n  =  0  by  §  29. 


Second,  suppose  that  the  proposed  construction  is  possible  with  ruler 
and  compasses.  The  straight  lines  and  circles  drawn  in  making  the  con- 
struction are  located  by  means  of  points  either  initially  given  or  obtained 
as  the  intersections  of  two  straight  lines,  a  straight  line  and  a  circle,  or 
two  circles.  Since  the  axes  of  coordinates  are  at  our  choice,  we  may 
assume  that  the  y-axis  is  not  parallel  to  any  of  the  straight  lines  employed 
in  the  construction.     Then  the  equation  of  any  one  of  our  lines  is 


(3) 


y  =  mx-\-b. 


Let  y  =  m'x-\-b'  be  the  equation  of  another  of  our  lines  which  inter- 
sects (3).     The  coordinates  of  their  point  of  intersection  are 


x  = 


b'-b 


m—m 


y- 


mb'  —  m'b 
m—m'  ' 


which  are  rational  functions  of  the  coefficients  of  the  equations  of  the 
two  lines. 

Suppose  that  a  line  (3)  intersects  the  circle 

(x-c)2+(y-d)2  =  r2, 

with  the  center  (c,  d)  and  radius  r.  To  find  the  coordinates  of  the  points 
of  intersection,  we  eliminate  y  between  the  equations  and  obtain  a  quad- 
ratic equation  for  x.    Thus  x  (and  hence  also  mx-\-b  or  y)  involves  no 


32  CONSTRUCTIONS  WITH  RULER  AND  COMPASSES         [Ch.  Ill 

irrationality  other  than  a  real  square  root,  besides  real  irrationalities 
present  in  m,  b,  c,  d,  r. 

Finally,  the  intersections  of  two  circles  are  given  by  the  intersections 
of  one  of  them  with  their  common  chord,  so  that  this  case  reduces  to  the 
preceding. 

.For  example,  a  side  of  a  regular  pentagon  inscribed  in  a  circle  of  radius  unity  is 
(Ex.  2  of  §  37)  

(4)  s  =  l\/lO-2V5, 

which  is  a  number  of  the  type  mentioned  in  the  criterion.  Hence  a  regular  pentagon 
can  be  constructed  by  ruler  and  compasses  (see  the  example  above  quoted) . 

31.  Cubic  Equations  with  a  Constructive  Root.  We  saw  that  the 
problem  of  the  duplication  of  a  cube  led  to  a  cubic  equation  (2).  We 
shall  later  show  that  each  of  the  problems,  to  trisect  an  angle,  and  to  con- 
struct regular  polygons  of  7  and  9  sides  with  ruler  and  compasses,  leads 
to  a  cubic  equation.  We  shall  be  in  a  position  to  treat  all  of  these  problems 
as  soon  as  we  have  proved  the  following  general  result. 

Theokem.  It  is  not  possible  to  construct  with  ruler  and  compasses  a 
line  whose  length  is  a  root  or  the  negative  of  a  root  of  a  cubic  equation  with 
rational  coefficients  having   no  rational  root. 

Suppose  that  x\  is  a  root  of 

(5)  x3+ax2-\-px+y  =  0      (a,  )3,  y  rational) 

such  that  a  line  of  length  xi  or  —  Xi  can  be  constructed  with  ruler  and  com- 
passes; we  shall  prove  that  one  of  the  roots  of  (5)  is  rational.  We  have 
only  to  discuss  the  case  in  which  xi  is  irrational. 

By  the  criterion  in  §  30,  since  the  given  numbers  in  this  problem  are 
a,  /3,  7,  all  rational,  x\  can  be  obtained  by  a  finite  number  of  rational 
operations  and  extractions  of  real  square  roots,  performed  upon  rational 
numbers  or  numbers  derived  from  them  by  such  operations.  Thus  x\ 
involves  one  or  more  real  square  roots,  but  no  further  irrationalities. 

As  in  the  case  of  (4),  there  may  be  superimposed  radicals.  Such  a 
two-story  radical  which  is  not  expressible  as  a  rational  function,  with 
rational  coefficients,  of  a  finite  number  of  square  roots  of  positive  rational 
numbers  is  said  to  be  a  radical  of  order  2.  In  general,  an  n-story  radical 
is  said  to  be  of  order  n  if  it  is  not  expressible  as  a  rational  function,  with 
rational  coefficients,  of  radicals  each  with  fewer  than  n  superimposed 
radicals,  the  innermost  ones  affecting  positive  rational  numbers. 


§31]  CUBIC  EQUATIONS  WITH  A  CONSTRUCTIBLE  ROOT  33 

We  agree  to  simplify  x\  by  making  all  possible  replacements  of  certain 
types  that  are  sufficiently  illustrated  by  the  following  numerical  examples. 

If  xi  involves  V3,  V5,  and  Vl5,  we  agree  to  replace  Vl5  by  VZ-VK 
If  xi  =  s  —  It,  where  s  is  given  by  (4)  and 


so  that  si=v5,  we  agree  to  write  x\  in  the  form  s  —  7\^5/s,  which  involves 
a  single  radical  of  order  2  and  no  new  radical  of  lower  order.  Finally, 
we  agree  to  replace  V 4  —  2V3  by  its  simpler  form  V3— 1. 

After  all  possible  simplifications  of  these  types  have  been  made,  the 
resulting  expressions  have  the  following  properties  (to  be  cited  as  our 
agreements):  no  one  of  the  radicals  of  highest  order  n  in  x\  is  equal  to 
a  rational  function,  with  rational  coefficients,  of  the  remaining  radicals 
of  order  n  and  the  radicals  of  lower  orders,  while  no  one  of  the  radicals 
of  order  n—  1  is  equal  to  a  rational  function  of  the  remaining  radicals  of 
order  n—\  and  the  radicals  of  lower  orders,  etc. 

Let  Vk  be  a  radical  of  highest  order  n  in  x\.     Then 

=  a+bVk 

Xl~c+dVW 

where  a,  b,  c,  d  do  not  involve  Vfc,  but  may  involve  other  radicals.  If 
d  =  0,  then  c^-0  and  we  write  e  for  a/c,  f  for  b/c,  and  get 

(6)  X!=e+fVh,  C/VO) 

where  neither  e  nor/  involves  Vk.  If  d^O,  we  derive  (6)  by  multiplying 
the  numerator  and  denominator  of  the  fraction  for  x\  by  c—dVk,  which 
is  not  zero  since  'Vk  =  c/d  would  contradict  our  above  agreements. 

By  hypothesis,  (6)  is  a  root  of  equation  (5).  After  expanding  the 
powers  and  replacing  the  square  of  V&  by  k,  we  see  that 

(7)  (e+/V^)3+«(e+/V^)2+/3(e+/V/7)+7  =  A+JBVfc, 

where  A  and  B  are  certain  polynomials  in  e,  f,  k  and  the  rational  numbers 
a,  j8,  7.  Thus  A+BVk  =  0.  If  B^O,  Vk=  -A/B  is  a  rational  function, 
with  rational  coefficients,  of  the  radicals,  other  than  v/b,  in  x\,  contrary 
to  our  agreements.     Hence  B  =  0  and  therefore  A  =  0. 

When  e—  /Vfc  is  substituted  for  x  in  the  cubic  function  (5),  the  result 


34  CONSTRUCTIONS  WITH  RULER  AND  COMPASSES         [Ch.  Ill 

is  the  left  member  of  (7)  with  V/b  replaced  by  —  V/c,  and  hence  the  result 
is  A  -  BVk.    But  A  =  B  =  0.     This  shows  that 

(8)  x2  =  e-fVk 

is  a  new  root  of  our  cubic  equation.  Since  the  sum  of  the  three  roots 
is  equal  to  —a  by  §  20,  the  third  root  is 

(9)  X3=  —  a— X\  —  X2=  —  a  —  2e. 

Now  a  is  rational.  If  also  e  is  rational,  xs  is  a  rational  root  and  we  have 
reached  our  goal.  We  next  make  the  assumption  that  e  is  irrational 
and  show  that  it  leads  to  a  contradiction.  Since  e  is  a  component  part 
of  the  constructible  root  (6),  its  only  irrationalities  are  square  roots. 
Let  Vs  be  one  of  the  radicals  of  highest  order  in  e.  By  the  argument 
which  led  to  (6),  we  may  write  e  =  e'+/'Vs,  whence,  by  (9), 

(9')  xz^g+hVs,  (h9*0) 

where  neither  g  nor  h  involves  Vs.  Then  by  the  argument  which  led 
to  (8),  g  —  AVs  is  a  root,  different  from  xz,  of  our  cubic  equation,  and  hence 
is  equal  to  x\  or  X2  since  there  are  only  three  roots  (§  16).     Thus 

g-hVs  =  e±fVk. 

By  definition,  Vs  is  one  of  the  radicals  occurring  in  e.  Also,  by  (9'), 
every  radical  occurring  in  g  or  h  occurs  in  x%  and  hence  in  e  =  %{— a— £3), 
by  (9),  a  being  rational.  Hence  V/c  is  expressible  rationally  in  terms 
of  the  remaining  radicals  occurring  in  e  and  /,  and  hence  in  x\,  whose  value 
is  given  by  (6).     But  this  contradicts  one  of  our  agreements. 

32.  Trisection  of  an  Angle.  For  a  given  angle  A,  we  can  construct 
with  ruler  and  compasses  a  line  of  length  cos  A  or  —cos  A,  namely  the 
adjacent  leg  of  a  right  triangle,  with  hypotenuse  unity,  formed  by  dropping 
a  perpendicular  from  a  point  in  one  side  of  A  to  the  other,  produced  if 
necessary.  If  it  were  possible  to  trisect  angle  A,  i.e.,  construct  the  angle 
A/3  with  ruler  and  compasses,  we  could  as  before  construct  a  line  whose 
length  is  ±cos  (A/3).  Hence  if  we  show  that  this  last  cannot  be  done 
when  the  only  given  geometric  elements  are  the  angle  A  and  a  line  of 
unit  length,  we  shall  have  proved  that  the  angle  A  cannot  be  trisected. 
We  shall  give  the  proof  for  A  =  120°. 

We  employ  the  trigonometric  identity 

cos  A  =4  cos3  -5 — 3  cos  -5-. 
o  o 


§  32]  TRISECTION  OF  AN  ANGLE  35 

Multiply  each  term  by  2  and  write  x  for  2  cos  (A/3).     Thus 

(10)  z3-3:r  =  2cosA. 
For  A  =  120°,  cos  A  =  —  £  and  (10)  becomes 

(11)  x?-Sx+l=0. 

Any  rational  root  is  an  integer  (§  27)  which  is  an  exact  divisor  of  the 
constant  term  (§24).     By  trial,  neither  +1  nor  —1  is  a  root.     Hence 

(11)  has  no  rational  root.     Hence  (§31)  it  is  not  possible  to  trisect  all 
angles  with  ruler  and  compasses. 

Certain  angles,  like  90°,  180°,  can  be  trisected.  When  A  =  180°,  the  equation 
(10)  becomes  x3— 3x=—  2  and  has  the  rational  root  x  =  l.  It  is  the  rationality  of  a 
root  which  accounts  for  the  possibility  of  trisecting  this  special  angle  180°. 

33.  Regular  Polygon  of  9  Sides,  Duplication  of  a  Cube.  Since  angle 
120°  cannot  be  trisected  with  ruler  and  compasses  (§  32),  angle  40°  can- 
not be  so  constructed  in  terms  of  angle  120°  and  the  line  of  unit  length 
as  the  given  geometric  elements.  Since  the  former  of  these  elements 
and  its  cosine  are  constructible  when  the  latter  is  given,  we  may  take 
the  line  of  unit  length  as  the  only  given  element.  In  a  regular  polygon 
of  9  sides,  the  angle  subtended  at  the  center  by  one  side  is  ^-360°  =  40°. 
Hence  a  regular  polygon  of  9  sides  cannot  be  constructed  with  ruler  and  com- 
passes. Here,  as  in  similar  subsequent  statements  where  the  given 
elements  are  not  specified,  the  only  such  element  is  the  line  of  unit  length. 

A  rational  root  of  x?  =  2  is  an  integer  (§  27)  which  is  an  exact  divisor 
of  2.  The  cubes  of  ±1  and  ±2  are  distinct  from  2.  Hence  there  is  no 
rational  root.  Hence  (§§  30,  31)  it  is  not  possible  to  duplicate  a  cube  with 
ruler  and  compasses. 

34.  Regular  Polygon  of  7  Sides.  If  we  could  construct  with  ruler 
and  compasses  an  angle  5  containing  360/7  degrees,  we  could  so  con- 
struct a  line  of  length  re  =  2  cos  5.  Since  7  5  =  360°,  cos3  5  =  cos4  5. 
But 

2  cos  SB  =  2(4  cos35  -  3  cos  5)  =  x*  -  3x, 

2  cos  45  =  2(2  cos2  25-1)  =4(2  cos25-l)2-2=  (z2-2)2-2. 
Hence 

0  =  x4  -  4x2 + 2  -  (x3  -  3x)  =  (x  -  2)  (x3  +x2  -  2x  - 1) . 

But  x  =  2  would  give  cos  5=1,  whereas  5  is  acute.     Hence 

(12)  x3+x2-2x-l=0. 


36  CONSTRUCTIONS  WITH  RULER  AND  COMPASSES         [Ch.  Ill 

Since  this  has  no  rational  root,  it  is  impossible  to  construct  a  regular 
polygon  of  7  sides  with  ruler  and  compasses. 

35.  Regular  Polygon  of  7  Sides  and  Roots  of  Unity.     If 

_  2tt,   .',    2tt 

R  =  cos-=-+t  siny, 

we  saw  in  §  10  that  R,  R2,  R3,  R4,  R5,  R6,  R7  =  l  give  all  the  roots  of  y7  =  1 
and  are  complex  numbers  represented  by  the  vertices  of  a  regular  polygon 
of  7  sides  inscribed  in  a  circle  of  radius  unity  and  center  at  the  origin  of 
coordinates.     By  §  6, 

1  2tt      .    .     2tt  d  .    1      o  2tt 

^  =  cos^ — tsm-y,         #+-^  =  2cosy. 

We  saw  in  §  34  that  2  cos  (27r/7)  is  one  of  the  roots  of  the  cubic 
equation  (12).  This  equation  can  be  derived  in  a  new  manner  by  utilizing 
the  preceding  remarks  on  7th  roots  of  unity.  Our  purpose  is  not  primarily 
to  derive  (12)  again,  but  to  illustrate  some  principles  necessary  in  the 
general  theory  of  the  construction  of  regular  polygons. 

Removing  from  y7  —  1  the  factor  y—1,  we  get 

(13)  y6+y»+t+yz+y2Jry+i=Q, 

whose  roots  are  R,  R2,  .  .  .  ,  R6.  Since  we  know  that  R-\-l/R  is  one  of 
the  roots  of  the  cubic  equation  (12),  it  is  a  natural  step  to  make  the  sub- 
stitution 

(14)  y+±  =  x 

in  (13).     After  dividing  its  terms  by  ys,  we  have 

(i3o  {y3+$+{y2+$+{y+l)+l=0- 

By  squaring  and  cubing  the  members  of  (14),  we  see  that 

(15)  </2+p  =  z2-2,         y3+±  =  v?-Sx. 

Substituting  these  values  in  (13;),  we  obtain 

(12)  x3+x2-2x-l=0. 

That  is,  the  substitution  (14)  converts  equation  (13)  into  (12). 


§  35]  REGULAR  POLYGON  OF  SEVEN  SIDES  3? 

If  in  (14)  we  assign  to  y  the  six  values  R,  .  .  .  ,  R6,  we  obtain  only 
three  distinct  values  of  x: 

(16)     X!=R-\-^=R-\-RQ,    X2  =  R2+^r2  =  R2+R5,     x3  =  R3+~  =  R^+R\ 

In  order  to  illustrate  a  general  method  of  the  theory  of  regular  poly- 
gons, we  start  with  the  preceding  sums  of  the  six  roots  in  pairs  and  find 
the  cubic  equation  having  these  sums  as  its  roots.  For  this  purpose  we 
need  to  calculate 

Xi+X2  +  X3,  XiX2+XiX3  +  X2X3,  X!X2X3. 

First,  by  (16), 

xi+x2+x3  =  R+R2+  .  .  .  +#6=-l, 

since  R,  .  .  .  ,  R6  are  the  roots  of  (13).     Similarly, 

x^+x^+xoXs  =  2(R+R2+  .  .  .  +RS)  =  -2, 

xix2x3  =  2+R+R2+  .  .  .  +#6  =  1. 

Consequently  (§  20),  the  cubic  having  xi,  x2,  x3  as  roots  is  (12). 

36.  Reciprocal  Equations.  Any  algebraic  equation  such  that  the 
reciprocal  of  each  root  is  itself  a  root  ot  the  same  multiplicity  is  called  a 
reciprocal  equation. 

The  equation  y1  —  1  =0  is  a  reciprocal  equation,  since  if  r  is  any  root,  1/r  is  evidently 
also  a  root.  Since  (13)  has  the  same  roots  as  this  equation,  with  the  exception  of  unity 
which  is  its  own  reciprocal,  (13)  is  also  a  reciprocal  equation. 

If  r  is  any  root  5^0  of  any  equation 

Ky)^t+  ■  •  •  +c=o, 

l/r  is  a  root  of  f(l/y)  =0  and  hence  of 


ynf(f)=i+  •••  +cyn=0. 


If  the  former  is  a  reciprocal  equation,  it  has  also  the  root  l/r,  so  that  every 
root  of  the  former  is  a  root  of  the  latter  equation.  Hence,  by  §  18,  the 
left  member  of  the  latter  is  indentical  with  cj(y).  Equating  the  constant 
terms,  we  have  c2  =  l,  c=±l.     Hence 

(17)  2/w/Q=±/(2/). 


38  CONSTRUCTIONS  WITH  RULER  AND  COMPASSES         [Ch.  Ill 

Thus  if  Piyn~%  is  a  term  of  f(y),  also  ±pty*  is  a  term.     Hence 
(18')  Ky)=yn±l+pi(yn-1±y)+p2(yn-2±y2)+  .... 

If  n  is  odd,  w  =  2£+l,  the  final  term  is  pt{ytJrl±yl),  and  ?/dbl  is  a  factor 
of /(?/).     In  view  of  (17),  the  quotient 

AW-® 

has  the  property  that 


»""iQQsQ(y)- 


Comparing  this  with   (17),  which  implied  (18'),  we  see  that  Q(y)=0  is 
a  reciprocal  equation  of  the  type 

(18)    y2t+l+c1(y2t-1t+y)+c2(y2t-2+y2)+  .  .  .  +c,_1(^+1+^-1)+^  =  0. 

If  n  is  even,  n  =  2£,  and  if  the  upper  sign  holds  in  (17),  then  (18')  is 
of  the  form  (18).  Next,  let  the  lower  sign  hold  in  (17).  Since  a  term 
ptyl  would  imply  a  term  —  pty\  we  have  pt  =  0.  The  final  term  in  (18') 
is  therefore  pt-i(yt+1  —  yt~1)-  Hence  f(y)  has  the  factor  y2—  1.  The 
quotient  q(y)  =f(y)  / \y2 '—  1)  has  the  property  that 


yn~\\^j=<i(y)- 


Comparing  this  with  (17)  as  before,  we  see  that  q(y)=0  is  of  the  form 
(18)  where  now  2t  =  n  —  2.  Hence,  at  least  after  removing  one  or  both 
of  the  factors  y±l,  any  reciprocal  equation  may  be  given  the  form  (18). 

The  method  by  which  (13)  was  reduced  to  a  cubic  equation  may  be 
used  to  reduce  any  equation  (18)  to  an  equation  in  x  of  half  the  degree. 
First,  we  divide  the  terms  of  (18)  by  yl  and  obtain 


(^)+4'-'+^)+  .-.+*..(»+§ 


+ct  =  0. 


Next,  we  perform  the  substitution  (14)  by  either  of  the  following  methods: 
We  may  make  use  of  the  relation 

to  compute  the  values  of  yt-\-l/yk  in  terms  of  x,  starting  with  the  special 


36] 


RECIPROCAL  EQUATIONS 


39 


cases  (14)  and  (15).     For  example, 

=  x(x3  -  Zx)  -  (x2  -  2)  =  x4  -  4rc2+2 

Or  we  may  employ  the  explicit  formula  (19)  of  §  107  for  the  sum  yt-\-l/yt 
of  the  kth  powers  of  the  roots  y  and  \/y  of  y2  —  xy+1  =0. 

37.  Regular  Polygon  of  9  Sides  and  Roots  of  Unity.     If 

R  =  cos  -q-+*  sin  -Q-, 

the  powers  R,  R2,  R4,  R5,  R7,  R8,  are  the  primitive  ninth  roots  of  unity 
(§11).     They  are  therefore  the  roots  of 

,9_! 


(19) 


y 


=2/6+2/3+l  =  0. 


Dividing  the  terms  of  this  reciprocal  equation  by  y3  and  applying  the  sec- 
ond relation  (15),  we  obtain  our  former  cubic  equation  (11). 


EXERCISES 

1.  Show  by  (16)  that  the  roots  of  (12)  are  2  cos  2tt/7,  2  cos  4tt/7,  2  cos  6tt/7. 

2.  The  imaginary  fifth  roots  of  unity  satisfy  y4+  B 
2/3+2/2+2/+l=0,  which  by  the   substitution   (14)  be- 
comes x2+x  —  1  =0.     It  has  the  root 


R+ 


R 


2w     1      /- 
=2  cos— =-(V5-l). 


In  a  circle  of  radius  unity  and  center  0  draw  two  per- 
pendicular diameters  AOA',  BOB'.  With  the  middle 
point  M  of  OA'  as  center  and  radius  MB  draw  a  circle 
cutting  OA  at  C  (Fig.  10).  Show  that  OC  and  BC 
are  the  sides  sw  and  s5  of  the  inscribed  regular  decagon 
and  pentagon  respectively.     Hints: 


MB- 


-W&, 


Sio  =  2  sin  18°  =  2  cos 


OC=|(V5-l), 

2* 


fiC  =  Vl+OC2  =  |VlO-2\/5, 


=  0(7, 


s62  =  (2sin36°)2  =  2M-cos  — j=-(10-2\/5),         §5  =  5(7. 


40  CONSTRUCTIONS  WITH  RULER  AND  COMPASSES         [Ch.  Ill 

3.  If  R  is  a  root  of  (19)  verify  as  at  the  end  of  §  35  that  R+R*,  R2+R7,  and  R*+R* 
are  the  roots  of  (11). 

4.  Hence  show  that  the  roots  of  (11)  are  2  cos  2ir/9,  2  cos  4tt/9,  2  cos  8tt/9. 

5.  Reduce  y11  =  1  to  an  equation  of  degree  5  in  x. 

6.  Solve  y5—  7y4+y3  —  y2+7y  —  1  =0  by  radicals.     [One  root  is  1.] 

7.  After  finding  so  easily  in  Chapter  I  the  trigonometric  forms  of  the  complex  roots 
of  unity,  why  do  we  now  go  to  so  much  additional  trouble  to  find  them  algebraically? 

8.  Prove  that  every  real  root  of  xi+ax2+b  =  0  can  be  constructed  with  ruler  and 
compasses,  given  lines  of  lengths  a  and  b. 

9.  Show  that  the  real  roots  of  x3  —  px  —  q  =  0  are  the  abscissas  of  the  intersections 
of  the  parabola  y=x2  and  the  circle  through  the  origin  with  the  center  (%q,  |+|p). 

Prove  that  it  is  impossible,  with  ruler  and  compasses : 

10.  To  construct  a  straight  line  representing  the  distance  from  the  circular  base 
of  a  hemisphere  to  the  parallel  plane  which  bisects  the  hemisphere. 

11.  To  construct  lines  representing  the  lengths  of  the  edges  of  an  existing  rectangular 
parallelopiped  having  a  diagonal  of  length  5,  surface  area  24,  and  volume  1,  2,  3,  or  5. 

12.  To  trisect  an  angle  whose  cosine  is  \,  I,  \,  \  or  p/q,  where  p  and  q  (q>l)  are 
integers  without  a  common  factor,  and  q  is  not  divisible  by  a  cube. 

Prove  algebraically  that  it  is  possible,  with  ruler  and  compasses: 

13.  To  trisect  an  angle  whose  cosine  is  (4a3  — 3ab2)/fe3,  where  the  integer  a  is  numer- 
ically less  than  the  integer  b;  for  example,  cos-1  ll/16ifa=— 1,  b=4. 

14.  To  construct  the  legs  of  a  right  triangle,  given  its  area  and  hypotenuse. 

15.  To  construct  the  third  side  of  a  triangle,  given  two  sides  and  its  area. 

16.  To  locate  the  point  P  on  the  side  BC  =  1  of  a  given  square  ABCD  such  that 
tne  straight  line  AP  cuts  DC  produced  at  a  point  Q  for  which  the  length  of  PQ  is  a  given 
number  g.  Show  that  y  =  BP  is  a  root  of  a  reciprocal  quartic  equation,  and  solve  it 
when  o  =  10, 

38.  The  Periods  of  Roots  of  Unity.  Before  taking  up  the  regular 
polygon  of  17  sides,  we  first  explain  another  method  of  finding  the  pairs 
of  imaginary  seventh  roots  of  unity  R  and  R6,  R2  and  R5,  R3  and  R4, 
employed  in  (16).  To  this  end  we  seek  a  positive  integer  g  such  that 
the  six  roots  can  be  arranged  in  the  order 

(20)  R,   R°,   Rg2,   Rg3,   Rgi,   R9\ 

where  each  term  is  the  gth  power  of  its  predecessor.  Trying  g  =  2,  we  find 
that  the  fourth  term  would  then  be  R8  =  R.  Hence  g^2.  Trying  #  =  3, 
we  obtain 

(21)  R,   R3,   R2,   R6,   R4,   R5, 
where  each  term  is  the  cube  of  its  predecessor. 


§  39]  REGULAR  POLYGON  OF  17  SIDES  41 

To  define  three  periods,  each  of  two  terms, 

(16')  R+R6,        R2+R5,        R3+R4, 

we  select  the  first  term  R  of  (21)  and  the  third  term  R6  after  it  and  add 
them,  then  the  second  term  it!3  and  the  third  term  R4  after  it,  and  finally 
R2  and  the  third  term  R5  after  it. 

We  may  also  define  two  periods,  each  of  three  terms, 

z1  =  R-{-R2-{-R4,    z2  =  Rz+R&+R5, 

by  taking  alternate  terms  in  (21). 

Since  Zi+Z2  =  — 1,  ZiZ2  =  3+R-\-  .  .  .  -\-R6  =  2,  zi  and  z2  are  the  roots  of  z2+z+2=0. 
Then  R,  R2,  R4  are  the  roots  of  w3—  ZiW2-\-z2w  — 1  =  0. 

39.  Regular  Polygon  of  17  Sides.     Let  R  be  a  root  ^1  of  x17  =  l. 
Then 

**^^=R16+R15+  .  .  .  +£+1=0. 
K —  1 

As  in  §  38,  we  may  take  g  =  3  and  arrange  the  roots  R,  .  .  . ,  R16  so  that 
each  is  the  cube  of  its  predecessor: 

R,  R3,R9,  R10,  R13,  R5,  R15,  Rn,  R16,  R14,  R8,  R7,  R4,  R12,  R2,  R6. 
Taking  alternate  terms,  we  get  the  two  periods,  each  of  eight  terms, 
yi=R+R9+Rl3+R15+R16+R8+R4-\-R2, 
y2  =  R3+R10+R5+Rn+R14+R7+R12+R6. 

Hence  yi+y2=  -1.    We  find  that  yiy2  =  4:  (R+  .  .  .  +Rl6)  =  -4.    Thus 

(22)  2/1,      y2      satisfy     y2+y-4:  =  0. 
Taking  alternate  terms  in  yi,  we  obtain  the  two  periods 

zi=JR+#13+#16+£4,         z2  =  Rg+R15+R8+R2. 
Taking  alternate  terms  in  y2)  we  get  the  two  periods 

Wl  =  R3+R*+Rl4+R12,       w2^R10+Rn+R7+R(i. 
Thus  z\ +z2  =  y\,  wi-\-w2  —  y2.     We  find  that  2i22  =  wiW2=  —1.     Hence 

(23)  z\,    22       satisfy      z2  —  y\Z  — 1  =  0, 

(24)  w\,    w2     satisfy     w2  —  y2w  —  1  =  0. 


42  CONSTRUCTIONS  WITH  RULER  AND  COMPASSES         [Ch.  Ill 

Taking  alternate  terms  in  z\,  we  obtain  the  periods 
Vl  =  R+R16,        v2  =  R13+R*. 
Now,  vi-\-V2  =  Zi,  viV2  =  wi.    Hence 

(25)  vi,    V2     satisfy     v2— ziv-\-wi=0, 

(26)  R,    R16     satisfy     P2-v1P+  1  =  0. 

Hence  we  can  find  R  by  solving  a  series  of  quadratic  equations.  Which 
of  the  sixteen  values  of  R  we  shall  thus  obtain  depends  upon  which  root 
of  (22)  is  called  y\  and  which  y2,  and  similarly  in  (23)-(26).  We  shall  now 
show  what  choice  is  to  be  made  in  each  such  case  in  order  that  we  shall 
finally  get  the  value  of  the  particular  root 

D  2t      .   .    2?r 

R  =  cos  yj+i  sin  t=. 

Then 

1  2tt     .  .    2tt  d    .  1       _        2tt 

-^  =  cos-=-ismj=,        vi  =  R  +p  =  2cos  y=, 

JR4  =  cos-y+«sm-w,        V2  =  /t4+Q4  =  2cos-=. 
Hence  vi  >  V2  >  0,  and  therefore  z\  =  yi+V2  >  0.     Similarly, 

tvj    i       1      i     7->r;    ,      1  «  67T    ,    _  lOx        _  07T        _  '7T        _ 

Wl  =#3+-^3+#5+^5  =  2  COS  Yy  +  2  COS  ^y-  =  2  COS  Jj"2  C°S  JJ>°» 

6-n-  IOt  12-n-  \4lT 

2/2  =  2  cos  T^+2  cos  Ty-+2  c°s  "77~+2  cos  TT      ' 

since  only  the  first  cosine  in  y%  is  positive  and  it  is  numerically  less  than 
the  third.     But  yiy2=  -4.     Hence  j/i>0.     Thus  (22)-(24)  give 

yi=h(VV7-i),  i/2  =  |(-Vl7-l), 


We  may  readily  construct  segments  of  these  lengths.  Evidently 
Vl7  is  the  length  of  the  hypotenuse  of  a  right  triangle  whose  legs  are  of 
lengths  1  and  4,  while  for  the  radical  in  z\  we  employ  legs  of  lengths  1 
and  \y\.     We  thus  obtain  segments  representing  the  coefficients  of  the 


40] 


REGULAR  POLYGONS 


43 


quadratic  equation  (25). 
larger  root  is 


Its  roots  may  be  constructed  as  in  §  29. 
2-k 


The 


v\  =  2  cos 


17' 


Hence  we  can  construct  angle  2-71-/17  with  ruler  and  compasses,  and  there- 
fore a  regular  polygon  of  17  sides. 

40.  Construction  of  a  Regular  Polygon  of  17  Sides.  In  a  circle  of 
radius  unity,  construct  two  per- 
pendicular diameters  AB,  CD, 
and  draw  tangents  at  A,  D, 
which  intersect  at  S  (Fig.  11). 
Find  the  point  E  in  AS  for  which 
AE  —  \AS,  by  means  of  two  bi- 
sections.    Then 

AE=h        OE  =  lVl7. 

Let  the  circle  with  center  E 
and  radius  OE  cut  AS  at  F  and 
F'.     Then 


F'    S 


AF  =  EF-EA  =  OE-\  =  \yi, 
AF'  =  EF'+EA  =  OE+\  =  -±y2, 

OF=VOA2+AF2  =  Vl+lVl2,        OF'  =  Vl+ly22. 

Let  the  circle  with  center  F  and  radius  FO  cut  AS  at  H,  outside  of  F'F; 
that  with  center  F'  and  radius  F'O  cut  AS  at  H'  between  F'  and  F.    Then 


AH=AF+FH=AF+0F=iyi+Vl+lyi2=zi, 
AH'=F'H'-F'A=OF'-AF'=wi. 

It  remains  to  construct  the  roots  of  equation  (25).  This  will  be  done 
as  in  §  29.  Draw  HTQ  parallel  to  AO  and  intersecting  OC  produced  at 
T.  Make  TQ  =  AH'.  Draw  a  circle  having  as  diameter  the  line  BQ 
joining  B—  (0,1)  with  Q=(zi,wi).  The  abscissas  ON  and  OM  of  the  inter- 
sections of  this  circle  with  the  rc-axis  OT  are  the  roots  of  (25).  Hence 
the  larger  root  v\  is  0M  =  2  cos(27r/17). 


44  CONSTRUCTIONS  WITH  RULER  AND  COMPASSES         [Ch.  in 

Let  the  perpendicular  bisector  LP  of  OM  cut  the  initial  circle  of  unit 
radius  at  P.     Then 

cos  LOP  =  0L  =  cos  ~,        LOP  =  ^. 

Hence  the  chord  CP  is  a  side  of  the  inscribed  regular  polygon  of  17 
sides,  constructed  with  ruler  and  compasses. 

41.  Regular  Polygon  of  n  Sides.  If  n  be  a  prime  such  that  n  —  1  is 
a  power  2h  of  2  (as  is  the  case  when  n  =  3,  5,  17),  the  n— 1  imaginary  nth 
roots  of  unity  can  be  separated  into  2  sets  each  of  2h~1  roots,  each  of  these 
sets  subdivided  into  2  sets  each  of  2h~2  roots,  etc.,  until  we  reach  the  pairs 
R,  1/R  and  R2,  l/R2,  etc.,  and  in  fact 1  in  such  a  manner  that  we  have  a 
series  of  quadratic  equations,  the  coefficients  of  any  one  of  which  depend 
only  upon  the  roots  of  quadratic  equations  preceding  it  in  the  series. 
Note  that  this  was  the  case  for  n  =  17  and  for  n  =  5.  It  is  in  this  manner 
that  it  can  be  proved  that  the  roots  of  xn  =  l  can  be  found  in  terms  of 
square  roots,  so  that  a  regular  polygon  of  n  sides  can  be  inscribed  by  ruler 
and  compasses,  provided  n  be  a  prime  of  the  form  2ft+l. 

If  n  be  a  product  of  distinct  primes  of  this  form,  or  2fc  times  such  a 
product  (for  example,  n  =  15,  30  or  6),  or  if  n  =  2m(m>  1),  it  follows  readily 
(see  Ex.  1  below)  that  we  can  inscribe  with  ruler  and  compasses  a  regular 
polygon  of  n  sides.     But  this  is  impossible  for  all  other  values  of  n. 

EXERCISES 

1.  If  os  and  b  are  relatively  prime  numbers,  so  that  their  greatest  common  divisor 
is  unity,  we  can  find  integers  c  and  d  such  that  ac-\-bd  =  \.  Show  that,  if  regular  poly- 
gons of  a  and  b  sides  can  be  constructed  and  hence  angles  2w/a  and  2-ir/b,  a  regular 
polygon  of  a-b  sides  can  be  derived. 

2.  If  p  =  2h+l  is  a  prime,  h  is  a  power  of  2.  For  h  =  2°,  21 ,  22,  23,  the  values  of  p 
are  3,  5,  17,  257  and  are  primes.  [Show  that  h  cannot  have  an  odd  factor  other  than 
unity.] 

3.  For  13th  roots  of  unity  find  the  least  g  (§38),  write  out  the  three  periods  each 
of  four  terms,  and  find  the  cubic  equation  having  them  as  roots. 

4.  For  the  primitive  ninth  roots  of  unity  find  the  least  g  and  write  out  the  three 
periods  each  of  two  terms. 

Solve  the  following  reciprocal  equations: 

5.  y4+4?/3-3?/2+%  +  l=0.  6.  y5-4y4+y3+y2-4y  +  l=0. 
7.  2y«-5y\+4yi-4y2+5y-2  =  0.         8.  y&  +  l  =31(?/  +  1)5. 

1  See  the  author's  article  "  Constructions  with  ruler  and  compasses;  regular  poly- 
gons," in  Monographs  on  Topics  of  Modern  Mathematics,  Longmans,  Green  and  Co., 
1911,  p.  374. 


CHAPTER  IV 

Solution  of  Cubic  and  Quartic  Equations;    Their  Discriminants 

42.  Reduced  Cubic  Equation.     If,  in  the  general  cubic  equation 

(1)  x3+bx2+cx+d  =  0, 

we  set  x  =  y—b/3,  we  obtain  the  reduced  cubic  equation 

(2)  y3+py+q  =  0, 
lacking  the  square  of  the  unknown  y,  where 

(3)  p  =  c-j,         9  =  d— 3+27- 

After  finding  the  roots  yi,  yz,  ys  of  (2),  we  shall  know  the  roots  of  (1): 

,..  b  b  b 

(4)  xi=yi—g,        x2  =  y2-y        x3  =  y3--. 

43.  Algebraic  Solution  of  the  Reduced  Cubic  Equation.  We  shall 
employ  the  method  which  is  essentially  the  same  as  that  given  by  Vieta 
in  1591.     We  make  the  substitution 

(5)  _„-.-£ 

in  (2)  and  obtain 

'■ff-0, 


27z3 
since  the  terms  in  z  cancel,  and  likewise  the  terms  in  1/z.     Thus 

(6)  .    z6+gz3-^  =  0. 


Solving  this  as  a  quadratic  equation  for  z3,  we  obtain 

2 


(7)  ,3=_|±VK,        B-.g)V8 


46  CUBIC  AND  QUARTIC  EQUATIONS  [Ch.  IV 

By  §  8,  any  number  has  three  cube  roots,  two  of  which  are  the  products 
of  the  remaining  one  by  the  imaginary  cube  roots  of  unity: 

(8)  w=-!+§Va;,     co2=-i-iV3;. 

We  can  choose  particular  cube  roots 


(9) 


=  ff^.       B^-|-VS, 


such  that  AB  =  —  p/3,  since  the  product  of  the  numbers  under  the  cube 
root  radicals  is  equal  to  (  —  p/3)3.     Hence  the  six  values  of  z  are 

A,      coA,      o>2A,      B,      uB,      u2B. 

These  can  be  paired  so  that  the  product  of  the  two  in  each  pair  is  —  p/3  • 

Hence  with  any  root  z  is  paired  a  root  equal  to  —p/(3z).     By  (5),  the  sum 
of  the  two  is  a  value  of  y.     Hence  the  three  values  of  y  are 

(10)  !/i=  A+B,        y2  =  o>A  +  u2B,        y3  =  u2A+a:B. 

It  is  easy  to  verify  that  these  numbers  are  actually  roots  of  (2),  For 
example,  since  co3  =  1,  the  cube  of  y2  is 

A3+B3+3uA2B+3co2AB2  =  -q-V{uA  +  u2B)  =  -q-py2, 

by  (9)  and  AB=-<p/3. 

The  numbers  (10)  are  known  as  Cardan's  formulas  for  the  roots  of  a 
reduced  cubic  equation  (2).  The  expression  A  +  B  for  a  root  was  first 
published  by  Cardan  in  his  Ars  Magna  of  1545,  although  he  had  obtained 
it  from  Tartaglia  under  promise  of  secrecy. 

Example  .     Solve  y%  —  15y  — 126  =  0 . 

Solution.    The  substitution  (5)  is  here  y  =  z+5/z.    We  get 

z6-126z3 +125  =  0,         z3  =  lorl25. 

The  pairs  of  values  of  z  whose  product  is  5  are  1  and  5,  w  and  5«2,  w2  and  5w.     Theii 
sums  6,  w+5w2,  and  a>2+5co  give  the  three  roots. 

EXERCISES 

Solve  the  equations: 

'1.  i/3-18?/ +35=0.  2.  x3+6x2+3a;+18=0. 

3.  y3-2y+4  =  0.  4.  28z3+9:e2-l=0. 


§  44]  DISCRIMINANT  OF  A  CUBIC  47 

44.  Discriminant.  The  product  of  the  squares  of  the  differences  of 
the  roots  of  any  equation  in  which  the  coefficient  of  the  highest  power  of 
the  unknown  is  unity  shall  be  called  the  discriminant  of  the  equation. 
For  the  reduced  cubic  (2),  the  discriminant  is 

(11)  (y  i  -  y2)2  (y  i  -  ys)2  (y2 -ys)2=-4:ps- 27  q2, 

a  result  which  should  be  memorized  in  view  of  its  important  applications. 
It  is  proved  by  means  of  (10)  and  co3  =  l,  co2+to  +  l  =0,  as  follows: 

yi-y2  =  (l-o>)(A-^B),       yi-yz  =  {l-^)(A-uB), 

2/2-2/3  =  (co-co2)  (A.-B), 

(l-co)(l-co2)=3,  co-co2  =  V3f. 

Since  1,  co,  co2  are  the  cube  roots  of  unity, 

(x  —  1)  (x  —  co)  (x  —  co2)  =  x3  —  1, 

identically  in  x.     Taking  x  =  A/B,  we  see  that 

(A  -B)(A-  co£)  (A  -  u2B)  =  A3-B3  =  2Vr, 

by  (9).     Hence 

(yi  -  2/2)  (yi  -  2/3)  (2/2  -  2/3)  =  6  VsVr  i. 

Squaring,  we  get  (11),  since   —  108it!=  —  4p3  —  27 q2  by   (7).       For  later 
use,  we  note  that  the  discriminant  of  the  reduced  cubic  is  equal  to  — 108  R. 
The  discriminant  A  of  the  general  cubic  (1)  is  equal  to  the  discriminant 
of  the  corresponding  reduced  cubic  (2).     For,  by  (4), 

xi-x2  =  yi-y2,        xi-xz  =  y1-y3,         X2-x3  =  y2-y3- 
Inserting  in  (11)  the  values  of  p  and  q  given  by  (3),  we  get 

(12)  A  =  18bcd-±b3d+b2c2-4:c3- 27  d2. 
It  is  sometimes  convenient  to  employ  a  cubic  equation 

(13)  ax3+bx2+cx+d  =  0      (a^O), 

in  which  the  coefficient  of  x3  has  not  been  made  unity  by  division.  The  product  P 
of  the  squares  of  the  differences  of  its  roots  is  evidently  derived  from  (12)  by  replacing 
b,  c,  d  by  b/a,  c/a,  d/a.     Hence 

(14)  a4P  =  18  abcd-4b3d+b2c2-<iac3-27a2d2. 
This  expression  (and  not  P  itself)  is  called  the  discriminant  of  (13). 


48  CUBIC  AND  QUARTIC  EQUATIONS  [Ch.  IV 

45.  Number  of  Real  Roots  of  a  Cubic  Equation.  A  cubic  equation 
with  real  coefficients  has  three  distinct  real  roots  if  its  discriminant  A  is  positive, 
a  single  real  root  and  two  conjugate  imaginary  roots  if  A  is  negative,  and  at 
least  two  equal  real  roots  if  A  is  zero. 

If  the  roots  %\,  X2,  X3  are  all  real  and  distinct,  the  square  of  the  differ- 
ence of  any  two  is  positive  and  hence  A  is  positive. 

If  xi  and  X2  are  conjugate  imaginaries  and  hence  £3  is  real  (§  21), 
{X1  —  X2)2  is  negative.  Since  x\—  23  and  X2  —  Xz  are  conjugate  imaginaries, 
their  product  is  positive.     Hence  A  is  negative. 

If  X\=X2,  A  is  zero.  If  X2  were  imaginary,  its  conjugate  would  be 
equal  to  xz  by  §  21,  and  X2,  X3  would  be  the  roots  of  a  real  quadratic 
equation.  The  remaining  factor  x  —  x\  of  the  cubic  would  have  real 
coefficients,  whereas  x\=x2  is  imaginary.  Hence  the  equal  roots  must 
be  real. 

Our  theorem  now  follows  from  these  three  results  by  formal  logic. 
For  example,  if  A  is  positive,  the  roots  are  all  real  and  distinct,  since 
otherwise  either  two  would  be  imaginary  and  A  would  be  negative,  or  two 
would  be  equal  and  A  would  be  zero. 

EXERCISES 

Compute  the  discriminant  and  find  the  number  of  real  roots  of 

Jl.  y3 -2^-4=0.  2.  y3-15y+4:  =  0. 

3.  y3-27y+54:  =  0.  4.  x3+4x2-llx+6  =  0. 

5.  Show  by  means  of  §  21  that  a  double  root  of  a  real  cubic  is  real. 

46.  Irreducible  Case.  When  the  roots  of  a  real  cubic  equation  are 
all  real  and  distinct,  the  discriminant  A  is  positive  and  i2=— A/108  is 
negative,  so  that  Cardan's  formulas  present  the  values  of  the  roots  in  a 
form  involving  cube  roots  of  imaginaries.  This  is  called  the  irreducible 
case  since  it  may  be  shown  that  a  cube  root  of  a  general  complex  number 
cannot  be  expressed  in  the  form  a+bi,  where  a  and  b  involve  only  real 
radicals.1  While  we  cannot  always  find  these  cube  roots  algebraically, 
we  have  learned  how  to  find  them  trigonometrically  (§8). 

Example.     Solve  the  cubic  equation  (2)  when  p=  — 12,  q=  —  8 V  2. 
Solution.     By  (7),  R=  —32.     Hence  formulas  (9)  become 

,4=V4\/2+4V/2i,         B  =  V  4V2-4V21. 
1  Author's  Elementary  Theory  of  Equations,  pp  35,  36. 


§  47]  TRIGONOMETRIC  SOLUTION  OF  A  CUBIC  49 

The  values  of  A  were  found  in  §  8.     The  values  of  B  are  evidently  the  conjugate  imagi- 
naries  of  the  values  of  A .     Hence  the  roots  are 

4  cos  15°,     4  cos  135°,     4  cos  255°. 


EXERCISES 

1.  Solve  y3-15y +4  =  0.  2.  Solve  y3-2y-l  =  0. 

/3.  Solve  y3 -ly +7  =  0.  4.  Solve  x3+3x2-2x -5=0. 

5.  Solve  x3+x2-2x-l=0.  ,6.  Solve  x3+Ax2 -7  =  0. 

47.  Trigonometric  Solution  of  a  Cubic  Equation  with  A>0.  When 
the  roots  of  a  real  cubic  equation  are  all  real,  i.e.,  if  R  is  negative,  they 
can  be  computed  simultaneously  by  means  of  a  table  of  cosines  with  much 
less  labor  than  required  by  Cardan's  formulas.  To  this  end  we  write 
the  trigonometric  identity 

cos  3A  =  4  cos3  A  —  3  cos  A 
in  the  form 

z3  —  \z  —  I  cos  3A=0        (2  =  cosA). 

In  the  given  cubic  y3+py-\-q  =  0  take  y  =  nz;  then 

*3  +  ^  +  4  =  0, 

which  will  be  identical  with  the  former  equation  in  z  if 


n  =  V-±p,     cos3A  =  -|g-=-\/-p3/27. 

Since  R  =  p3/27-\-q2/4  is  negative,  p  must  be  negative,  so  that  n  is  real 
and  the  value  of  cos  2>A  is  real  and  numerically  less  than  unity.  Hence 
we  can  find  3A  from  a  table  of  cosines.     The  three  values  of  z  are  then 

cos  A,         cos(A  +  120°),         cos(A+240°). 

Multiplying  these  by  n,  we  obtain  the  three  roots  y  correct  to  a  number 
of  decimal  places  which  depends  on  the  tables  used. 

EXERCISES 

/l.  For  ^-2^-1=0,  show  that  n2  =  8/3,  cos  3^1  =  V27/32  3A=23°17' 0", 
cos  A=  0.99084,  cos  (A  +120°)  =  -0.61237,  cos  (A  +240°)  =  -0,37847,  and  that  the 
roots  y  are  1.61804,  -1,  -0.61804. 

2.  Solve  Exs.  1,  3,  4,  5,  6  of  §  46  by  trigonometry. 


50  CUBIC  AND  QUARTIC  EQUATIONS  [Ch.  IV 

48.  Ferrari's  Solution  of  the  Quartic  Equation.     The  general  quartic 
equation 

(15)  x4+bx3+ex2+dx+e  =  0, 

or  equation  of  degree  four,  becomes  after  transposition  of  terms 

rc4+foc3=  —  ex2— dx— e. 

The  left  member  contains  two  of  the  terms  of  the  square  of  x2+\bx. 
Hence  by  completing  the  square,  we  get 

(x2 + %bz)2  =  {\b2-  c)x2  -dx-e. 

Adding  (x2jr%bx)y-\-\y2  to  each  member,  we  obtain 

(16)  (x2+±bx+hy)2=(lb2-c+y)x2+(hby-d)x+ly2-e. 

The  second  member  is  a  perfect  square  of  a  linear  function  of  x  if  and 
only  if  its  discriminant  is  zero  (§  12): 

(hby-d)2-±£b2-c+y)  (ly2-e)=0, 

which  may  be  written  in  the  form 

(17)  y3  -cy2+(bd-4:e)y-b2e+4ce-d2  =  0. 

Choose  any  root  y  of  this  resolvent  cubic  equation  (17).     Then  the 
right  member  of  (16)  is  the  square  of  a  linear  function,  say  mx-\-n.     Thus 

(18)  x2jr\bx-\-\y  =  mx-\-n    or    x2jr\bx-\-\y= —mx  —  n. 

The  roots  of  these  quadratic  equations  are  the  four  roots  of  (16)  and 
hence  of  the  equivalent  equation  (15).  This  method  of  solution  is  due 
to  Ferrari  (1522-1565). 

Example.     Solve  x4+2x3-12a;2-10:c+3=0. 

Solution.    Here  b  =  2,  c=  — 12,  d=  — 10,  e  =  3.     Hence  (17)  becomes 

y3  +  12y2-32y-  256  =  0, 
which  by  Ex.  2  of  §  24  has  the  root  y  =  —4.     Our  quartic  may  be  written  in  the  form 

(,x1+x)*  =  13x*  +  10z-3. 
Adding  (x2+x)  (—4) +4  to  each  member,  we  get 

(x2+x-2y  =  Qx2+Gx  +  l  =  (3x+l)2, 

z2+x-2  =  ±(3x+l),         x2-2x-3  =  0ora;2+4z-l=0, 

whose  roots  are  3,  —1,  — 2±  V  5.     As  a  check,  note  that  the  sum  of  the  roots  is  —2. 


§  50]  DISCRIMINANT  OF  A  QUARTIC  51 

EXERCISES 

■1.  Solve  x4-8x3+9x24-8x- 10  =  0.     Note  that  (17)  is  (y-9)  (y2-24)=0. 

2.  Solve  x4-2x3-7x2+8x+12  =  0.  Since  the  right  member  of  (16)  is  (S+y)  (x2-x) 
+b2-12,  use  7/= -8. 

3.  Solve  x4-3x2+6x-2  =  0. 

4.  Solve  x4-2x2-8x -3=0.  ^5.  Solve  x4-10x2-20x -16  =  0. 

49.  Roots  of  the  Resolvent  Cubic  Equation.  Let  2/1  be  the  root  y 
which  was  employed  in  §  48.  Let  X\  and  X2  be  the  roots  of  the  first 
quadratic  equation  (18),  and  x%  and  x±  the  roots  of  the  second.     Then 

xix2  =  %yi-n,        X3Xi  =  iyi+n,        xiX2+X3X±=yi. 

If,  instead  of  y\,  another  root  2/2  or  2/3  of  the  resolvent  cubic  (17)  had  been 
employed  in  §  48,  quadratic  equations  different  from  (18)  would  have 
been  obtained,  such,  however,  that  their  four  roots  are  x\,  x2,  X3,  £4,  paired 
in  a  new  manner.  The  root  which  is  paired  with  x\  is  x2  or  X3  or  x±.  It 
is  now  plausible  that  the  values  of  the  three  y's  are 

(19)  yi=XiX2+X3X4;,  y2=XiX3+X2X4,  y3=XiX4:+X2X3. 

To  give  a  more  formal  proof  that  the  y's  given  by  (19)  are  the  roots 
of  (17),  we  employ  (§  20) 

Xi+X2-\-X3+X±=  —b,  XiX2X3+XiX2X±  +  XiX3Xi-\-X2X3X±=  ~d, 

X\X2-\-XiX3-\-XiX±-\-X2X3-{-X2X4-\-X3X4L  =  C,  X\X2X3X±  =  e. 

From  these  four  relations  we  conclude  that 

2/i+2/2+2/3  =  c, 

2/12/2+2/12/3  +  2/21/3=  (Xi+X2  +  X3+X±)(XiX2X3-\r  .  .  .   +X2X3X4)  —  ±XXX2X3X± 

=  bd  —  4e, 

yiy2y3=(XlX2X3+   .  .  .  )2-\-XiX2X3X4i{{Xi+   .  .  .  )2~4:(XiX2+   .  .  .  )} 

=  d2+e(62-4c). 
Hence  (§  20)  2/1,  2/2,  2/3  are  the  roots  of  the  cubic  equation  (17). 

50.  Discriminant.  The  discriminant  A  of  the  quartic  equation 
(15)  is  defined  to  be  the  product  of  the  squares  of  the  differences  of  its  roots: 

A=(Xi-X2)2(Xi-X3)2(Xi-X4)2(x2-X3)2(x2-X±)2(X3-X4)2. 


52  CUBIC  AND  QUARTIC  EQUATIONS  [Ch.  IV 

The  fact  that  A  is  equal  to  the  discriminant  of  the  resolvent  cubic 
equation  (17)  follows  at  once  from  (19),  by  which 

2/i  —  2/2  =  (xi—x±)  (x2—xz),       yi  —  y3  =  (xi-X3)(x2-X4:), 

y2-y3  =  (xi-x2)(x3-x±),      (yi-y2)2(yi-y3)2(y2-y3)2  =  &. 

Hence  (§44)  A  is  equal  to  the  discriminant  —  4p3  —  27 q2  of  the  reduced 
cubic  F3+pF+g  =  0,  obtained  from  (17)  by  setting  y=Y-\-c/S.     Thus 

(20)  y  =  bd-4e-\c2,         q=  -tfe+lbcd+^ce-d2-^. 

Theorem.  The  discriminant  of  any  quartic  equation  (15)  is  equal  to 
the  discriminant  of  its  resolvent  cubic  equation  and  therefore  is  equal  to  the 
discriminant  —  4p3  —  27q2  of  the  corresponding  reduced  cubic  Yz-\-pY-\-q  =  0, 
whose  coefficients  have  the  values  (20). 

EXERCISES 

1.  Find  the  discriminant  of  a;4  — 3xi+x2+3x  — 2  =  0  and  show  that  the  equation 
has  a  multiple  root. 

2.  Show  by  its  discriminant  that  x4— 8x3+22x2—  24z+9  =  0  has  a  multiple  root. 

3.  If  a  real  quartic  equation  has  two  pairs  of  conjugate  imaginary  roots,  show  that 
its  discriminant  A  is  positive.  Hence  prove  that,  if  A<0,  there  are  exactly  two  real 
roots. 

4.  Hence  show  that  xi— 3x3-\-3x2— Sx+2  =  0  has  two  real  and  two  imaginary  roots. 

51.  Descartes'  Solution  of  the  Quartic  Equation.  Replacing  x  by 
2  —  6/4  in  the  general  quartic  (15),  we  obtain  the  reduced  quartic  equation 

(21)  z4+qz2+rz+s  =  0, 

lacking  the  term  with  z8.     We  shall  prove  that  we  can  express  the  left 
member  of  (21)  as  the  product  of  two  quadratic  factors 

(z2+2kz+l)(z2-2kz+m)=z4+(l+m-4:k2)z2+2k(m-l)z+lm. 

The  conditions  are 

l-\-m  —  4k2  — q,         2k(m  —  l)=r,         lm  =  s. 

If  k^O,  the  first  two  give 

2l  =  q+4k2-^,         2m  =  q+4k2+^. 

Inserting  these  values  in  2l-2m  =  4s,  we  obtain 

(22)  64/c6+32gfc4+4(g2-4s)/c2-r2  =  0. 


§  51]  DESCARTES'  SOLUTION  OF  A  QUARTIC  53 

The  latter  may  be  solved  as  a  cubic  equation  for  k2.  Any  root  fc2^0 
gives  a  pair  of  quadratic  factors  of  (21) : 

(23)  z2±2kz+±q+2k2^F^. 

The  four  roots  of  these  two  quadratic  functions  are  the  four  roots  of  (21). 
This  method  of  Descartes  (1596-1650)  therefore  succeeds  unless  every 
root  of  (22)  is  zero,  whence  q  =  s  =  r  =  0,  so  that  (18.)  is  the  trivial  equation 

24  =  0.  *! 

For  example,  consider  z4— Sz2+6z  — 2  =  0.     Then  (22)  becomes 

64&6-3-32/c4+4-17A;2-36  =  0. 

The  value  k2  =  l  gives  the  factors  z2-\-2z  —  1,  z2  — 2z-\-2.  Equating  these  to  zero,  we 
find   the  four  roots  -1±V2,  1±V-1. 

52.  Symmetrical  Form  of  Descartes'  Solution.     To  obtain  this  sym- 
metrical form,  we  use  all  three  roots  k\2,  k22,  k32  of  (22).     Then 

k{2+k22+kf  =  ~k,  h2k22ks2=^. 

It  is  at  our  choice  as  to  which  square  root  of  k\2  is  denoted  by  -\-k\  and 
which  by  —ki,  and  likewise  as  to  ±k2,  ±&3-  For  our  purposes  any 
choice  of  these  signs  is  suitable  provided  the  choice  give 

(24)  fci*afcs=-g. 

Let  fci^O.     The  quadratic  function  (23)  is  zero  for  k  =  k\  if 

(z±k1r=-^-k12±~=k22+k32^8J^^=(k2-Fk3)2. 

Hence  the  four  roots  of  the  quartic  equation  (21)  are 

(25)  ki+k2+k3j         ki  —  k2  —  k3,         —ki+k2  —  kz,         —ki  —  k2+k3. 

EXERCISES 

1.  Solve  Exs.  4,  5  of  §  48  by  the  method  of  Descartes. 

2.  By  writing  yh  y2,  ys  for  the  roots  h2,  k22,  k32  of 

(26)  64?/3+32gt/2+4(g2-4s)7/-r2  =  0, 
show  that  the  four  roots  of  (21)  are  the  values  of 

(27)  z  =  Vy~1  +  Vy2  +  Vy~i 


54  CUBIC  AND  QUARTIC  EQUATIONS  [Ch.  IV 

for  all  combinations  of  the  square  roots  for  which 

(28)  Vyl-V^-Vy^-1-. 


3.  Euler  (1707-1783)  solved  (21)  by  assuming  that  it  has  a  root  of  the  form  (27). 
Square  (27),  transpose  the  terms  free  of  radicals,  square  again,  replace  the  last  factor 
of  8 Vy1y2y3  (v  2/i+ v  y2+ v  2/s)  by  z,  and  identify  the  resulting  quartic  in  2  with  (21). 
Show  that  j/i,  2/2,  2/3  are  the  roots  of  (26)  and  that  relation  (28)  holds. 

4.  Find  the  six  differences  of  the  roots  (25)  and  verify  that  the  discriminant  A  of 
(21)  is  equal  to  the  quotient  of  the  discriminant  of  (26)  by  46. 

5.  In  the  theory  of  the  inflexion  points  of  a  plane  cubic  curve  there  occurs  the 
equation 

zi-Sz2-^Tz—^S2  =  0. 


Show  that  (26)  now  becomes 


»-f '-«  HW-(I 


and  that  the  roots  of  the  quartic  equation  are 


±  y/\s + <fc  ±  y/\s + oj^c  ±  V±,s + o*  Vc, 

where  u  is  an  imaginary  cube  root  of  unity  and  the  signs  are  to  be  chosen  so  that  the 
product  of  the  three  summands  is  equal  to  -\-\T. 

MISCELLANEOUS  EXERCISES 

/-l.  Find  the  coordinates  of  the  single  real  point  of  intersection  of  the  parabola 
y  =  x2  and  the  hyperbola  xy— 4x+2/+6  =  0. 

2.  Show  that  the  abscissas  of  the  points  of  intersection  of  y  =  x2  and  ax2—xy-\-y2  — 
x  —  (a +5) y  —  6  =  0  are  the  roots  of  x4— x3  —  5x2— x— 6  =  0.  Compute  the  discriminant 
of  the  latter  and  show  that  only  two  of  the  four  points  of  intersection  are  real. 

3.  Find  the  coordinates  of  the  two  real  points  in  Ex.  2. 

4.  A  right  prism  of  height  h  has  a  square  base  whose  side  is  b  and  whose  diagonal 
is  therefore  b V  2.  If  v  denotes  the  volume  and  d  a  diagonal  of  the  prism,  v  =  hb2  and 
d2  =  A2+(6 V  2)2.  Multiply  the  last  equation  by  h  and  replace  hb2  by  v.  Hence  h3—d2h 
+2i>  =  0.     Its  discriminant  is  zero  if  d  =  3\  3,  i'  =  27;    find  h. 

5.  Find  the  admissible  values  of  h  in  Ex.  4  when  d  =  12,  v  =  332.5. 

6.  Find  a  necessary  and  sufficient  condition  that  quartic  equation  (15)  shall  have 
one  root  the  negative  of  another  root. 

Hint:   (xi+x2)  (x3+x4)  =q  —  yi-     Hence  substitute  q  for  y  in  (17). 

7.  In  the  study  of  parabolic  orbits  occurs  the  equation  tan  2^  +  3  tan3  ^v  =  t. 
Prove  that  there  is  a  single  real  root  and  that  it  has  the  same  sign  as  t. 

8.  In  the  problem  of  three  astronomical  bodies  occurs  the  equation  x3+az+2  =  0. 
Prove  that  it  has  three  real  roots  if  and  only  if  a'S.  —3. 


CHAPTER  V 


The  Graph  of  an  Equation 

53.  Use  of  Graphs  in  the  Theory  of  Equations.  To  find  geometrically 
the  real  roots  of  a  real  equation  f(x)  =  0,  we  construct  a  graph  of  y  =f(x) 
and  measure  the  distances  from  the  origin  0  to  the  intersections  of  the 
graph  and  the  rc-axis,  whose  equation  is  y  =  0. 

For  example  to  find  geometrically  the  real 
roots  of 


(1) 


6x-3  =  0, 


we  equate  the  left  member  to  y  and  make  a 
graph  of 


(10 


y  =  x2  —  6#  —  3. 


ClM 

Y 

f(7,a 

0 

Q         /      „ 

\ 

4 

(0,-3): 

11 

*(6r3> 

0,-8) 

i(5,-8) 

( 

2- 

11)\ 

p; 

1X4,-11) 

We  obtain  the  parabola  in  Fig.  12.      Of  the 

points  shown,  P  has   the   abscissa   x  =  OQ  =  4: 

and  the  ordinate  y=—  QP=— 11.     From  the 

points  of  intersection  of  t/  =  0  (the  rr-axis  OX) 

with  the  parabola,  we  obtain  the  approximate  (3.-12) 

values  6.46  and  -0.46  of  the  roots  of  (1).  Fig.  12 

EXERCISES 

/\.  Find  graphically  the  real  roots  of  x2— 6x+7  =  0. 
Hint:   For  each  x,  y  =  x2—6x+7  exceeds  the  y  in  (1')  by  10,  so  that  the  new  graph 
is  obtained  by  shifting  the  parabola  in  Fig.  12  upward  10  units,  leaving  the  axes  OX 
and  OY  unchanged.     What  amounts  to  the  same  thing,  but  is  simpler  to  do,  we  leave 
the  parabola  and  OY  unchanged,  and  move  the  axis  OX  downward  10  units. 

2.  Discuss  graphically  the  reality  of  the  roots  of  x2  —  6x  +  12  =  0. 

3.  Find  graphically  the  roots  of  x2  —  62 +9  =  0. 

54.  Caution  in  Plotting.     If  the  example  set  were 

(2)  y  =  %x±-Ux*-§x2+Ux-2, 

one  might  use  successive  integral  values  of  x,  obtain  the  points  (  —  2,  180), 

55 


56 


GRAPHS 


[Ch.  V 


(-1,  0),  (0,  -2),  (1,  -6),  (2,  0),  (3,  220),  all  but  the 
first  and  last  of  which  are  shown  (by  crosses)  in  Fig. 
13,  and  be  tempted  to  conclude  that  the  graph  is  a 
U-shaped  curve  approximately  like  that  in  Fig.  12 
and  that  there  are  just  two  real  roots,  —  1  and  2,  of 

(2')  8x4-14a;3-9x2+llx-2-0. 

But  both  of  these  conclusions  would  be  false.  In  fact, 
the  graph  is  a  W-shaped  curve  (Fig.  13)  and  the 
additional  real  roots  are  \  and  ^. 

This  example  shows  that  it  is  often  necessary  to 
employ  also  values  of  x  which  are  not  integers.  The 
purpose  of  the  example  was,  however,  not  to  point 
out  this  obvious  fact,  but  rather  to  emphasize  the 
chance  of  serious  error  in  sketching  a  curve  through  a 
number  of  points,  however  numerous.  The  true  curve 
between  two  points  below  the  z-axis  may  not  cross  the 
z-axis,  or  may  have  a  peak  and  actually  cross  the  z-axis 
twice,  or  may  be  an  M-shaped  curve  crossing  it  four 
times,  etc. 

For  example,  the  graph  (Fig.  14)  of 

(3)  y  =  £3+4z2-ll 

crosses  the  z-axis  only  once;  but  this  fact  can- 
not be  established  by  a  graph  located  by  a  num- 
ber of  points,  however  numerous,  whose  abscissas 
are  chosen  at  random. 

We  shall  find  that  correct  conclusions  regard- 
ing the  number  of  real  roots  may  be  deduced 
from  a  graph  whose  bend  points  (§  55)  have  been 
located. 

55.  Bend  Points.     A  point  (like  M  or  M'  in 

Fig.  14)  is  called  a  bend  point   of   the   graph  of 

y  =f(x)  if  the  tangent  to  the  graph  at  that  point 

is  horizontal  and  if  all  of  the  adjacent  points  of 

the  graph  lie  below  the  tangent  or  all  above  the  _ 

.  Fig.  14 

tangent.     The  first,   but  not  the  second,  condi- 


Fig.  13 


55] 


BEND  POINTS 


57 


tion  is  satisfied  by  the  point  0  of  the  graph  of  y  =  xs  given  in  Fig.  15  (see 
§  57).  In  the  language  of  the  calculus,  f(x)  has  a  (relative)  maximum  or 
minimum  value  at  the  abscissa  of  a  bend  point  on  the  graph  of  y=f(x). 


Fig.  15 


Fig.  16 


Let  P=(x,y)  and  Q  =  (x-\-h,  Y)  be  two  points  on  the  graph,  sketched 
in  Fig.  16,  of  y—f{x).  By  the  slope  of  a  straight  line  is  meant  the  tangent 
of  the  angle  between  the  line  and  the  rc-axis,  measured  counter-clockwise 
from  the  latter.     In  Fig.  16,  the  slope  of  the  straight  line  PQ  is 


(4) 


Y-y     f(x+h)-f(x) 


h 


h 


For  equation  (3) ,  f(x)  =  x3 +4z2  —  11.     Hence 
f(x+h)  =  (x+hf+4;(x+K)2- 11 

=  rc3+4z2-ll  +  (3z2+8x)  ^+(3x4-4)  h2+h* 

The  slope  (4)  of  the  secant  PQ  is  therefore  here 

3z2+8z+(3z+4)/i+/i2. 

Now  let  the  point  Q  move  along  the  graph  toward  P.  Then  In  approaches 
the  value  zero  and  the  secant  PQ  approaches  the  tangent  at  P.  The 
slope  of  the  tangent  at  P  is  therefore  the  corresponding  limit  3z2+8z 
of  the  preceding  expression.     We  call  3a:2+8:r  the  derivative  of  x3+4x2—  1 1 . 


58  GRAPHS  [Ch.  V 

In  particular,  if  P  is  a  bend  point,  the  slope  of  the  (horizontal)  tangent 
at  P  is  zero,  whence  3x2-{-8x  =  0,  x  =  0  or  x=—  f.  Equation  (3)  gives 
the  corresponding  values  of  y.     The  resulting  points 

M=  (0,-11),        M'  =  (--|,-f^ 

are  easily  shown  to  be  bend  points.  Indeed,  for  :r.>0  and  for  x  between 
—  4  and  0,  x2  (x+4)  is  positive,  and  hence  f(x)>  —11  for  such  values  of 
x,  so  that  the  function  (3)  has  a  relative  minimum  at  x  =  0.  Similarly, 
there  is  a  relative  maximum  at  x=  —  f.  We  may  also  employ  the  general 
method  of  §  59  to  show  that  M  and  M'  are  bend  points.  Since  these  bend 
points  are  both  below  the  x-axis  we  are  now  certain  that  the  graph 
crosses  the  x-axis  only  once. 

The  use  of  the  bend  points  insures  greater  accuracy  to  the  graph  than 
the  use  of  dozens  of  points  whose  abscissas  are  taken  at  random. 

56.  Derivatives.  We  shall  now  find  the  slope  of  the  tangent  to  the 
graph  of  y=f(x),  where /(re)  is  any  polynomial 

(5)  f(x)=a0xn+a1xn-1+  .  .  .  -f-On-iS+a,,. 

We  need  the  expansion  of  f(x-\-h)  in  powers  of  x.  By  the  binomial 
theorem, 

a0(x+h)n  =  aoxn     +na0xn-1h         +n^~1)a0,gn-2ft2  +  .  .  .  , 

ai(x+h)n-1=a1xn-1  +  (n-l)a1xn-2h+(™~1)!;n~2\ixn-3h2+  .  .  .  , 


an_2(x+h)2  =  an_2x2jr2an_2xh  +aB_2/i2, 

aa-1(x+h)=an_1x  +an^1h, 

The  sum  of  the  left  members  is  evidently  f(x-\-h).  On  the  right,  the 
sum  of  the  first  terms  (i.e.,  those  free  of  h)  is  f(x).  The  sum  of  the  coef- 
ficients of  h  is  denoted  by  f(x),  the  sum  of  the  coefficients  of  \ln?  is  denoted 
by /"(#),  •  •  •  j  the  sum  of  the  coefficients  of 

hk 
1-2  ...  k 


§  56]  DERIVATIVES,  TAYLOR'S  THEOREM  59 

is  denoted  by  /(i)  (x) .     Thus 

(6)  f(x)=na0xn-1  +  (n-l)a1xn-2+  .  .  .  +2aw_2rc+oM_l5 

(7)  f"(x)=n(n-l)aoxn-2+(n-l)  (n-2)arxn-3+  .  .  .  +2an_2, 
etc.     Hence  we  have 

(8)    Kx+h)=f(x)+f(x)h+r(^+r"(x 


1-2  ,J     K  yl-2-3 

hn 
r!  n! 


+  .    ,  +/«(*£+  •  •  •  +/(ra)(*S 


where  r!  is  the  symbol,  read  r  factorial,  for  the  product  1-2-3  .  .  .  (r—l)r. 
Here  r  is  a  positive  integer,  but  we  include  the  case  r  =  0  by  the  definition, 
0!=1. 

This  formula  (8)  is  known  as  Taylor's  theorem  for  the  present  case  of 
a  polynomial /(a;)  of  degree  n.  We  call  fix)  the  (first)  derivative  of  fix), 
and  f"(x)  the  second  derivative  of  f(x),  etc.  Concerning  the  fact  that 
f"{x)  is  equal  to  the  first  derivative  of  f'(x)  and  that,  in  general,  the  kih 
derivative  fa)(x)  of  f(x)  is  equal  to  the  first  derivative  of  /(*_1)(a:)>  see 
Exs.  6-9  of  the  next  set. 

In  view  of  (8),  the  limit  of  (4)  as  h  approaches  zero  is  f'(x).  Hence 
f{x)  is  the  slope  of  the  tangent  to  the  graph  of  y=f(x)  at  the  point  (x,  y). 

In  (5)  and  (6),  let  every  a  be  zero  except  ao.  Thus  the  derivative  of 
aoxn  is  naoXn~l,  and  hence  is  obtained  by  multiplying  the  given  term  by 
its  exponent  n  and  then  diminishing  its  exponent  by  unity.  For  example, 
the  derivative  of  2x3  is  Qx2. 

Moreover,  the  derivative  of  f(x)  is  equal  to  the  sum  of  the  derivatives 

of  its  separate  terms.     Thus  the  derivative  of  x3  +  4z2  — 11  is  3x2-\-8x, 

as  found  also  in  §  55. 

EXERCISES 

1.  Show  that  the  slope  of  the  tangent  to  y  =  8x3  —  22x2  +  13x—2  at  (x,  y)  is  24x2- 
44x+13,  and  that  the  bend  points  are  (0.37,  0.203),  (1.46,  —5.03),  approximately. 
Draw  the  graph. 

2.  Prove  that  the  bend  points  of  y  =  x3-2z-5  are  (.82,  -6.09),  (-.82,  -3.91), 
approximately.     Draw  the  graph  and  locate  the  real  roots. 

3.  Find  the  bend  points  of  y  =  x3+6z2+8x+8.     Locate  the  real  roots. 

4.  Locate  the  real  roots  of  f(x)  =x4+x3— x  —  2=0. 

Hints:  The  abscissas  of  the  bend  points  are  the  roots  of  /'(x)  =4x3+3x2  — 1  =0. 
The  bend  points  of  y=f'(x)  are  (0,  —  1)  and  (— \,  —  f ),  so  that  ?{£)  =0  has  a  single  real 
root  (it  is  just  less  than  \).  The  single  bend  point  of  y=f(x)  is  (|,  —  f-g-),  approxi- 
mately. 


60  GRAPHS  [Ch.  V 

5.  Locate  the  real  roots  of  x6— 7xi— 3x2+7  =  0. 

6.  Prove  that/"(x),  given  by  (7),  is  equal  to  the  first  derivative  of  f'(x). 

7.  If  f(x)  =fi(x) +fo(x) ,  prove  that  the  kth.  derivative  of  /  is  equal  to  the  sum  of 
the  kth  derivatives  of /i  and/2.     Use  (8). 

8.  Prove  that  /(S)0)  is  equal  to  the  first  derivative  of  f(k~1)(x).     Hint:  prove  this 
for  f  =  axm;  then  prove  that  it  is  true  for  /  =/i +/2  if  true  for/i  and/2. 

9.  Find  the  third  derivative  of  x6+5x4  by  forming  successive  first  derivatives; 
also  that  of  2x5  —  7x3+x. 

10.  Prove  that  if  g  and  k  are  polynomials  in  x,the  derivative  of  gk  is  g'k+gk'.  Hint: 
multiply  the  members  of  g(x+h)=g(x)+g'(x)h+  .  .  .and  k(x+h)  =k(x)+k'(x)h+  .  .  . 
and  use  (8)  for  f=gk. 

57.  Horizontal  Tangents.  If  (x,  y)  is  a  bend  point  of  the  graph  of 
y=f(x),  then,  by  definition,  the  slope  of  the  tangent  at  (x,  y)  is  zero. 
Hence  (§56),  the  abscissa  a;  is  a  root  of  f(x)=0.  In  Exs.  1-5  of  the 
preceding  set,  it  was  true  that,  conversely,  any  real  root  of  f(x)=Q  is 
the  abscissa  of  a  bend  point.  However,  this  is  not  always  the  case. 
We  shall  now  consider  in  detail  an  example  illustrating  this  fact.  The 
example  is  the  one  merely  mentioned  in  §  55  to  indicate  the  need  of  the 
second  requirement  made  in  our  definition  of  a  bend  point. 

The  graph  (Fig.  15)  of  y  =  x?  has  no  bend  point  since  x3  increases  when 
x  increases.  Nevertheless,  the  derivative  3a;2  of  x3  is  zero  for  the  real 
value  x  =  0.  The  tangent  to  the  curve  at  (0,  0)  is  the  horizontal  lrae 
y  =  0.  It  may  be  thought  of  as  the  limiting  position  of  a  secant  through 
0  which  meets  the  curve  in  two  further  points,  seen  to  be  equidistant 
from  0.  When  one,  and  hence  also  the  other,  of  the  latter  points  approaches 
0,  the  secant  approaches  the  position  of  tangency.  In  this  sense  the 
tangent  at  0  is  said  to  meet  the  curve  in  three  coincident  points,  their 
abscissas  being  the  three  coinciding  roots  of  x3  =  0.  In  the  language  of 
§  17,  x3  =  0  has  the  triple  root  x  =  0.  The  subject  of  bend  points,  to  which 
we  recur  in  §  59,  has  thus  led  us  to  a  digression  on  the  important  subject 
of  multiple  roots. 

58.  Multiple  Roots.     In  (8)  replace  x  by  a,  and  h  by  x-a.     Then 
(9)  ftx)  =/(«)+/Xa)(^-«)+/^«)^^2+////(«)^^3+  .  .  . 


(ra— 1)!  m! 

By  definition  (§  17)  a  is  a  root  of /(re)  =0  of  multiplicity  m  ii  fix)  is  exactly 


§  58]  MULTIPLE  ROOTS  61 

divisible  by  (re  — a)m,  but  not  by  (re  —  a)m+1.     Hence  a  is  a  root  of  multi- 
plicity m  of  /(re)  =0  if  and  only  if 

(10)     f{a)  =  0,       f(a)=0,        f"(a)=0,   ...,       /<—»(«)  =0,        fm\a)^0. 

For  example,  xi-\-2x3  =  0  has  the  triple  root  x  =  0  since  0  is  a  root,  and  since  the 
first  and  second  derivatives  4x3+6x2  and  12x2-)-12x  are  zero  for  x  =  0,  while  the  third 
derivative  24x+12  is  not  zero  for  x  =  0. 

If  in  (9)  we  replace  /  by  /'  and  hence  /(4)  by  fa+1),  or  if  we  differentiate 
every  term  with  respect  to  x,  we  see  by  either  method  that 

(x-a)m-2 


(to- 2)! 


(11)  ff(x)  =/'(«)+/"(«)(*-«)  +  .  .  .  +/<-» 

^      w  (to-1)!  ^ 

Let /(re)  and/'(rr)  have  the  common  factor  (re— a)m~1,  but  not  the  com- 
mon factor  (x—a)m,  where  m>l.  Since  (11)  has  the  factor  (x-a)m_1,  we 
have  f'(a)=0,  .  .  .  ,  f(m~1)(a)=0.  Since  also  /(re)  has  the  factor  x  —  a, 
evidently  /(a)=0.  Then,  by  (9),  fix)  has  the  factor  (x-a)m,  which, 
by  hypothesis,  is  not  also  a  factor  of  /'(re).  Hence,  in  (11),  /(to)(q;)^0. 
Thus,  by  (10),  a  is  a  root  of  /(re)  =0  of  multiplicity  to. 

Conversely,  let  a  be  a  root  of  /(re)  =  0  of  multiplicity  to.  Then  rela- 
tions (10)  hold,  and  hence,  by  (11),  /'(re)  is  divisible  by  (x—a)"1-1,  but 
not  by  (re— a)m.  Thus /(re)  and /'(re)  have  the  common  factor  (re— a)"1-1, 
but  not  the  common  factor  (re— a)m. 

We  have  now  proved  the  following  useful  result. 

Theorem.  If  /(re)  and  f'(x)  have  a  greatest  common  divisor  g(x) 
involving  x,  a  root  of  gf(rc)=0  of  multiplicity  to— 1  is  a  root  of  /(rc)=0  of 
multiplicity  to,  and  conversely  any  root  of  /(re)  =  0  of  multiplicity  m  is  a  root 
°f  g(x)  =  Q  °f  multiplicity  to—  1. 

In  view  of  this  theorem,  the  problem  of  finding  all  the  multiple  roots 
of  /(re)  =  0  and  the  multiplicity  of  each  multiple  root  is  reduced  to  the 
problem  of  finding  the  roots  of  g(x)  =0  and  the  multiplicity  of  each. 

For  example,  let/(x)  =x3— 2x2— 4x+8.     Then 

fix)  =3x2-4x-4,         9/(.r)  =/'(x)(3x-2)  -32(x-2). 

Since  x  —  2  is  a  factor  of  fix),  it  may  be  taken  to  be  the  greatest  common  divisor  of  fix) 
and  fix),  the  choice  of  the  constant  factor  c  in  c(x— 2)  being  here  immaterial.  Hence 
2  is  a  double  root  of  /(x)  =0,  while  the  remaining  root  —2  is  a  simple  root. 


62  GRAPHS  [Ch.  V 

EXERCISES 

1.  Prove  that  x3  —  7x2+15x  —  9  =  0  has  a  double  root. 

2.  Show  that  x4  —  8x2+16  =  0  has  two  double  roots. 

3.  Prove  that  x4  —  6x2  — 8x  — 3  =  0  has  a  triple  root. 

4.  Test  x4-8x3+22x2-24x+9  =  0  for  multiple  roots. 

5.  Test  x3—  6x2+llx  —  6  =  0  for  multiple  roots. 

6.  Test  x4-9x3+9x2+81x- 162=0  for  multiple  roots. 

59.  Ordinary  and  Inflexion  Tangents.  The  equation  of  the  straight 
line  through  the  point  (a,  /3)  with  the  slope  s  is  y  —  (3  =  s(x  — a).  The  slope 
of  the  tangent  to  the  graph  of  y=f(x)  at  the  point  (a,  /3)  on  it  is  s  =/'(«) 
by  §  56.     Also,  /3  =/(«).     Hence  the  equation  of  the  tangent  is 

(12)  y=f(a)+f'(a)(x-a). 

By  subtracting  the  members  of  this  equation  from  the  corresponding 
members  of  equation  (9),  we  see  that  the  abscissas  x  of  the  points  of  inter- 
section of  the  graph  of  y  =f(x)  with  its  tangent  satisfy  the  equation 

J   w     2,     -rj    w     3!     -i-....-rv         w  (m-l)! 


^      v  y     m! 


=  0. 


Here  the  term  containing /(m_1)  (a)  must  evidently  be  suppressed  if  ra  =  2, 
since  the  term  containing  f(m)  (a)  then  coincides  with  the  first  term. 

If  a:  is  a  root  of  multiplicity  m  of  this  equation,  i.e.,  if  the  left  member 
is  divisible  by  (x—a)m,  but  not  by  (x—a)m+1,  the  point  (a,  /3)  is  counted 
as  m  coincident  points  of  intersection  of  the  curve  with  its  tangent  (just 
as  in  the  case  of  y  =  x3  and  its  tangent  y  =  0  in  §  57).  This  will  be  the  case 
if  and  only  if 

(13)        f"(a)=0,         /'"(a)=0,  .  .  .  ,  /(— D(a)=0,  fm)(a)^0, 

in  which  m>l  and,  as  explained  above,   only  the  final  relation  f"(a)?^0 
is  retained  if  m  =  2.     If  ra  =  3,  the  conditions  are  f'(a)  =0,  f(3)(a)^0. 

For  example,  if  /(x)  =x4  and  a=0,  then  /"(0)  =/"'  (0)  =0,  /c4)(0)  =  24^0,  so  that 
m  =  4.  The  graph  of  y  =  x4  is  a  [/-shaped  curve,  whose  intersection  with  the  tangent 
(the  x-axis)  at  (0,  0)  is  counted  as  four  coincident  points  of  intersection. 

Given  f(x)  and  a,  we  can  find,  as  in  the  preceding  example,  the  value 
of  m  for  which  relations  (13)  hold.     We  then  apply  the 


59] 


ORDINARY  AND  INFLEXION  TANGENTS 


63 


Theorem.  If  m  is  even  (m>0),  the  points  of  the  curve  in  the  vicinity 
of  the  point  of  tangency  (a,  /3)  are  all  on  the  same  side  of  the  tangent,  which 
is  then  called  an  ordinary  tangent.  But  if  m  is  odd  (ra>  1),  the  curve  crosses 
the  tangent  at  the  point  of  tangency  (a,  /3),  and  this  point  is  called  an  inflexion 
point,  while  the  tangent  is  called  an  inflexion  tangent. 

For  example,  in  Fig.  15,  OX  is  an  inflexion  tangent,  while  the  tangent  at  any  point 
except  0  is  an  ordinary  tangent.  In  Figs.  18,  19,  20,  the  tangents  at  the  points  marked 
by  crosses  are  ordinary  tangents,  but  the  tangent  at  the  point  midway  between  them 
and  on  the  y-axis  is  an  inflexion  tangent. 

To  simplify  the  proof,  we  first  take  as  new  axes  lines  parallel  to  the 
old  axes  and  intersecting  at  (a,  /3).  In  other  words,  we  set  x  — a  =  X, 
y  —  j3=Y,  where  X,  Y  are  the  coordinates  of  {x,  y)  referred  to  the  new 
axes.  Since  j3  =/(«),  the  tangent  (12)  becomes  Y=f(a)X,  while,  by  (9), 
y=f(x)=fiJrf'(a)  (x—a)-\-  .  .  .  becomes 


(14)  Y=f(a)X+f"(*)¥-+ 


ml 


iX,Y) 


after  omitting  terms  which  are  zero  by  (13). 
To  simplify  further  the  algebraic  work, 
we  pass  to  oblique  axes,1  the  new  ?/-axis 
coinciding  with  the  F-axis,  while  the  new 
z-axis  is  the  tangent,  the  angle  between 
which  ajad  the  X-axis  is  designated  by  d. 
Then 

tan  6=f'(a). 
By  Fig.  17, 

X  =  xcos6>,         Y-y=f'{a)X. 

Hence  when  expressed  in  terms  of  the 
new  coordinates  x,  y,  the  tangent  is  y  =  0,  while  the  equation  (14)  of  the 
curve  becomes 

fm)(a)  cos" 


y=cxm+dxm+1  + 


9*0. 


For  x  sufficiently  small  numerically,  whether  positive  or  negative, 
the  sum  of  the  terms  after  cxm  is  insignificant  in  comparison  with  cxm, 

1  Since  the  earlier  x,  y  do  not  occur  in  (14)  and  the  new  equation  of  the  tangent, 
we  shall  designate  the  final  coordinates  by  x,  y  without  confusion. 


64  GRAPHS  [Ch.  V 

so  that  y  has  the  same  sign  as  cxm  (§  64).  Hence,  if  m  is  even,  the  points 
of  the  curve  in  the  vicinity  of  the  origin  and  on  both  sides  of  it  are  all 
on  the  same  side  of  the  x-axis,  i.e.,  the  tangent.  But,  if  m  is  odd,  the  points 
with  small  positive  abscissas  x  lie  on  one  side  of  the  rc-axis  and  those  with 
numerically  small  negative  abscissas  lie  on  the  opposite  side. 

Our  transformations  of  coordinates  changed  the  equations  of  the 
curve  and  of  its  tangent,  but  did  not  change  the  curve  itself  and  its  tangent. 
Hence  our  theorem  is  proved. 

By  our  theorem,  a  is  the  abscissa  of  an  inflexion  point  of  the  graph 
of  y=f(x)  if  and  only  if  conditions  (13)  hold  with  m  odd  (m>l).  These 
conditions  include  neither  f(a)=0  nor  f'(a)=0,  in  contrast  with  (10). 
In  the  theory  of  equations  we  are  primarily  interested  in  the  abscissas 
a  of  only  those  points  of  inflexion  whose  inflexion  tangents  are  horizontal, 
and  are  interested  in  them,  because  we  must  exclude  such  roots  a  of 
f{x)  =0  when  seeking  the  abscissas  of  bend  points,  which  are  the  important 
points  for  our  purposes.  A  point  on  the  graph  at  which  the  tangent  is 
both  horizontal  and  an  ordinary  tangent  is  a  bend  point  by  the  definition 
in  §55.  Hence  if  we  apply  our  theorem  to  the  special  case  f'(a)=0, 
we  obtain  the  following 

Criterion.  Any  root  a  of  f'(x)=0  is  the  abscissa  of  a  bend  point  of 
the  graph  of  y  =f(x)  or  of  a  point  with  a  horizontal  inflexion  tangent  according 
as  the  value  of  m  for  which  relations  (13)  hold  is  even  or  odd. 

For  example,  if  f(x)  =xi,  then  a  =  0  and  to =4,  so  that  (0,  0)  is  a  bend  point  of  the 
[/-shaped  graph  of  y  =  xi.  If  f(x)=x3,  then  a  =  0  and  to  =  3,  so  that  (0,  0)  is  a  point 
with  a  horizontal  inflexion  tangent  (OX  in  Fig.  15)  of  the  graph  of  y  =  x3. 

EXERCISES 

1.  If  /(x)=3x5+5z3+4,  the  only  real  root  of  f'(x)=0  is  x  =  0.  Show  that  (0,  4) 
is  an  inflexion  point,  and  thus  that  there  is  no  bend  point  and  hence  that  f(x)  =  0  has  a 
single  real  root. 

2.  Prove  that  x3—  3x2-\-3x-\-c  =  0  has  an  inflexion  point,  but  no  bend  point. 

3.  Show  that  xb  —  ICte3— 20x2  —  15x+c  =  0  has  two  bend  points  and  no  horizontal 
inflexion  tangents. 

4.  Prove  that  3x5— 40rr3+240x+c  =  0  has  no  bend  point,  but  has  two  horizontal 
inflexion  tangents. 

5.  Prove  that  any  function  x3—3ax2+  ...  of  the  third  degree  can  be  written  in 
the  form/(x)  =  (x—  a)3+ax+b.  The  straight  line  having  the  equation  y  =  ax-\rb  meets 
the  graph  of  y  =f{x)  in  three  coincident  points  with  the  abscissa  a  and  hence  is  an 
inflexion  tangent.  If  we  take  new  axes  of  coordinates  parallel  to  the  old  and  inter- 
secting at  the  new  oriein  (a,  0),  i.e.,  if  we  make  the  transformation  x  =  X+a,  y=?Y, 


60] 


REAL  ROOTS  OF  A  CUBIC 


65 


of  coordinates,  we  see  that  the  equation  /(x)=0  becomes  a  1  educed  cubic  equation 
X3+pX+g=0(§42). 

6.  Find  the  inflexion    tangent  to   y=x3+Qx2  — 3x+l   and   transform    x3+Qx2  — 3x 
+  1=0  into  a  reduced  cubic  equation. 

60.  Real  Roots  of  a  Real  Cubic  Equation.     It  suffices  to  consider 

f{x)=x3-?>lx+q  (1^0), 

in  view  of  Ex.  5  above.     Then  /'  =  3  (x2-l),  f"  =  6x.     If  l<0,  there  is 
no  bend  point  and  the  cubic  equation  f(x)  =0  has  a  single  real  root. 
If  l>0,  there  are  two  bend  points 

(VT,      q-2lVl),       (-VT,      q+2lVT), 

which  are  shown  by  crosses  in  Figs.  18-20  for  the  graph  of  y=f(x)  in  the 


-2Z  vT-=g-=2Z  /I 
Fig.  20 


three  possible  cases  specified  by  the  inequalities  shown  below  the  figures. 
For  a  large  positive  x,  the  term  x3  in  f(x)  predominates,  so  that  the  graph 
contains  a  point  high  up  in  the  first  quad- 
rant, thence  extends  downward  to  the 
right-hand  bend  point,  then  ascends  to 
the  left-hand  bend  point,  and  finally  de- 
scends. As  a  check,  the  graph  contains 
a  point  far  down  in  the  third  quadrant, 
since  for  x  negative,  but  sufficiently  large 
numerically,  the  term  a;3  predominates  and  the  sign  of  y  is  negative. 

If  the  equality  sign  holds  in  Fig.  18  or  Fig.  19,  a  necessary  and  sufficient 
condition  for  which  is  q2  =  4l3,  one  of  the  bend  points  is  on  the  z-axis,  and 
the  cubic  equation  has  a  double  root.  The  inequalities  in  Fig.  20  hold 
if  and  only  if  q2<4l3,  which  implies  that  7>0.  Hence  x3  —  3lx-\-q  =  Q 
has  three  distinct  real  roots  if  and  only  if  q2<4l3,  a  single  real  root  if  and 
only  if  q2~>4l3,  a  double  root  (necessarily  real)  if  and  only  if  q2  —  4l3  and  1^0, 
and  a  triple  root  if  q2  =  4l3  —  0. 


66  GRAPHS  [Ch.  V 

EXERCISES 

Find  the  bend  points,  sketch  the  graph,  and  find  the  number  of  real  roots  of 
/L.  x3+2z-4  =  0.  2.  x3-7x+7=0. 

3.  z3-2x-l=0.  4.  x3+Qx2-dx+l=0. 

5.  Prove  that  the  inflexion  point  of  y  =  x3  —  3lx+q  is  (0,  q). 

6.  Show  that  the  theorem  in  the  text  is  equivalent  to  that  in  §  45. 

7.  Prove  that,  if  m  and  n  are  positive  odd  integers  and  m>n,  xm+pxn+q  =  0  has 
no  bend  point  and  hence  has  a  single  real  root  if  p>0;  but,  if  p<0,  it  has  just  two 
bend  points  which  are  on  the  same  side  or  opposite  sides  of  the  z-axis  according  as 

np\m      I   nq 
m  /         \m—'i 

is  positive  or  negative,  so  that  the  number  of  real  roots  is  1  or  3  in  the  respective  cases. 

8 .  Draw  the  graph  oiy  =  xi—x2.  By  finding  its  intersections  with  the  line  y  =  mx + b, 
solve  x4—  x2— mx  —  6  =  0. 

9.  Prove  that,  if  p  and  q  are  positive,  x2m  —  px2n+q  =  0  has  four  distinct  real  roots, 
two  pairs  of  equal  roots,  or  no  real  root,  according  as 

np\m     (    nq   \m~n     n 

J         >0,       =0,      or       <0. 
</ 

10.  Prove  that  no  straight  line  crosses  the  graph  of  y=f(x)  in  more  than  n  points  if 
the  degree  n  of  the  real  polynomial /(x)  exceeds  unity.  [Apply  §  16.]  This  fact  serves 
as  a  check  on  the  accuracy  of  a  graph. 

61.  Definition  of  Continuity  of  a  Polynomial.  Hitherto  we  have 
located  certain  points  of  the  graph  of  y=f(x),  where  f(x)  is  a  polynomial 
in  x  with  real  coefficients,  and  taken  the  liberty  to  join  them  by  a  con- 
tinuous curve. 

A  polynomial  f(x)  with  real  coefficients  shall  be  called  continuous  at 
x  =  a,  where  a  is  a  real  constant,  if  the  difference 

D=f(a+h)-f(a) 

is  numerically  less  than  any  assigned  positive  number  p  for  all  real  values 
of  h  sufficiently  small  numerically. 

62.  Any  Polynomial  f(x)  with  real  Coefficients  is  continuous  at  x  =  a, 
where  a  is  any  real  Constant.     Taylor's  formula  (8)  gives 

D=mh+£f§V+  ■  •  •  +1f2.(a]nhn' 

This  polynomial  is  a  special  case  of 

F  =  aih-\-a2h?+  .  .  .  +anhn. 


63] 


ROOT  LOCATED  BY  CHANGE  OF  SIGN 


67 


We  shall  prove  that,  if  a\,  .  .  .  ,  an  are  all  real,  F  is  numerically  less  than 
any  assigned  positive  number  p  for  all  real  values  of  h  sufficiently  small 
numerically.  Denote  by  g  the  greatest  numerical  value  of  a\,  .  .  .  ,  an. 
If  h  is  numerically  less  than  k,  where  k<l,  we  see  that  F  is  numerically 
less  than 

g(k+k2+  .  .  .  +k»)<g-^-i<p,         if  k<-£-. 

Hence  a  real  polynomial  f(x)  is  continuous  at  every  real  value  of  x.  But 
the  function  tan  x  is  not  continuous  at  x  =  90°  (§  63). 

63.  Root  between  a  and  b  if  f(a)  and  f(b)  have  opposite  Signs.     // 

the  coefficients  of  a  polynomial  fix)  are  real  and  if  a  and  b  are  real  numbers 
such  that  f(a)  and  f(b)  have  opposite  signs,  the  equation  f{x)  =  0  has  at  least 
one  real  root  between  a  and  b;  in  fact,  an  odd  number  of  such  roots,  if  an 
m-fold  root  is  counted  as  m  roots. 

The  only  argument 1  given  here  (other  than  that  in  Ex.  5  below)  is 
one  based  upon  geometrical  intuition.     We  are  stating  that,  if  the  points 

(a,   /(a)),         (b,  f(b)) 

lie  on  opposite  sides  of  the  z-axis,  the  graph  of  y=f{x)  crosses  the  rc-axis 
once,  or  an  odd  number  of  times,  between  the  vertical  lines  through 
these  two  points.  Indeed,  the  part  of  the  graph  be- 
tween these  verticals  is  a  continuous  curve  having  one 
and  only  one  point  on  each  intermediate  vertical  line, 
since  the  function  has  a  single  value  for  each  value 
of  x. 

This  would  not  follow  for  the  graph  of  y2  =  x,  which 
is  a  parabola  with  the  z-axis  as  its  axis.  It  may  not 
cross  the  z-axis  between  the  two  initial  vertical  lines, 
but  cross  at  a  point  to  the  left  of  each. 

A  like  theorem  does  not  hold  for  f{x)  =  tan  x,  when 
x  is  measured  in  radians  and  0<a<7r/2<6<x,  since 
tan  x  is  not  continuous  at  x  =  t/2.  When  t  increases 
from  a  to  ir/2,  tan  x  increases  without  limit.  When 
x  decreases  from  b  to  x/2,  tan  x  decreases  without 
limit.     There  is  no  root  between  a  and  b  of  tan  x  =  0. 


a    2L 


Fig.  21 


1  An  arithmetical  proof  based  upon  a  refined  theory  of  irrational  numbers  is  given 
in  Weber's  Lehrbuch  der  Algebra,  ed.  2,  vol.  1,  p.  123. 


68  GRAPHS  [Ch.  V 

EXERCISES 

1.  Prove  that  8x3— 4x2  —  18x+9=0  has  a  root  between  0  and  1,  one  between  1  and 
2,  and  one  between  —2  and  —1. 

2.  Prove  that  16x4— 24x2  +  16a;— 3=0  has  a  triple  root  between  0  and  1,  and  a 
simple  root  between  —2  and  —1. 

3.  Prove  that  if  a<b<c  .  .  .  <l,  and  a,  /3,  .  .  .  ,  X  are  positive,  these  quantities 
being  all  real, 

a  0  y  X 

+  -^r+-^+  ■  •  •  + -+t=0 

x—a     x—o    x—c  x—L 

has  a  real  root  between  a  and  b,  one  between  b  and  c,  .  .  .  .  one  between  k  and  I,  and 
if  t  is  negative  one  greater  than  I,  but  if  t  is  positive  one  less  than  a. 

4.  Verify  that  the  equation  in  Ex.  3  has  no  imaginary  root  by  substituting  r+si 
and  r  —  si  in  turn  for  x,  and  subtracting  the  results. 

5.  Admitting  that  an  equation  f(x)  =xn-\-  .  .  .  =0  with  real  coefficients  has  n  roots, 
show  algebraically  that  there  is  a  real  root  between  a  and  b  if  f(a)  and  f(b)  have  opposite 
signs.    Note  that  a  pair  of  conjugate  imaginary  roots  c±di  are  the  roots  of 

(x-c)2+d2  =  0 

and  that  this  quadratic  function  is  positive  if  x  is  real.  Hence  if  xi,  .  .  . ,  xr  are  the 
real  roots  and 

<t>(x)=(x—  Xi)  .  .  .  (X  —  Xr), 

then  <t>(a)  and  <f>(b)  have  opposite  signs.  Thus  a—xt  and  b—x%  have  opposite  signs  for 
at  least  one  real  root  x%.     (Lagrange.) 

64.  Sign  of  a  Polynomial.     Given  a  polynomial 

f{x)=a0xn+a1xn-x+  ...  +an  (a0^0) 

with  real  coefficients,  we  can  find  a  positive  number  P  such  that  f(x)  has 
the  same  sign  as  oqx71  when  x>P.     In  fact, 

/Cr)=z»(ao+4>),        0-^+*+...+* 

By  the  result  in  §  62,  the  numerical  value  of  <j>  is  less  than  that  of  a® 
when  1/x  is  positive  and  less  than  a  sufficiently  small  positive  number, 
say  1/P,  and  hence  when  x>P.  Then  ao+<£  has  the  same  sign  as  ao, 
and  hence  f{x)  the  same  sign  as  aoXn. 

The  last  result  holds  also  when  rr  is  a  negative  number  sufficiently  large 
numerically.  For,  if  we  set  x=—  X,  the  former  case  shows  that  /(  —  X) 
has  the  same  sign  as  (—  l)naoXn  when  X  is  a  sufficiently  large  positive 
number. 


§  65]  ROLLE'S  THEOREM  69 

We  shall  therefore  say  briefly  that,  for  re  =  +<*>,  f(x)  has  the  same 
sign  as  ao;  while,  for  x—  —  oo,  f(x)  has  the  same  sign  as  ao  if  n  is  even, 
but  the  sign  opposite  to  ao  if  n  is  odd. 

EXERCISES 

1.  Prove  that  xz-\-axi-\-bx— 4  =  0  has  a  positive  real  root  [use  a:  =  0  and  x  =  +  co  ]. 

2.  Prove  that  x3+ax2+bx+4i  =  0  has  a  negative  real  root  [use  x  =  0  and  x  =  —  co  ]. 

3.  Prove  that  if  a0>  0  and  n  is  odd,  a0xn+  .  .  .  +an  =  0  has  a  real  root  of  sign  opposite 
to  the  sign  of  an  [use  x  =  —  oo  ,  0,  +  oo  ]. 

4.  Prove  that  xi+ax3+bx2+cx— 4  =  0  has  a  positive  and  a  negative  root. 

5.  Show  that  any  equation  of  even  degree  n  in  which  the  coefficient  of  xn  and  the 
constant  term  are  of  opposite  signs  has  a  positive  and  a  negative  root. 

65.  Rolle's  Theorem.  Between  two  consecutive  real  roots  a  and  b  of  f(x) 
=  0,  there  is  an  odd  number  of  real  roots  of  fix)  =  0,  a  root  of  multiplicity  m 
being  counted  as  m  roots. 

Let 

f(x)  =  (x-a)T(x-b)sQ(x),  a<b, 

where  Q(x)  is  a  polynomial  divisible  by  neither  x  —  a  nor  x—b.  Then 
by  the  rule  for  the  derivative  of  a  product  (§  56,  Ex.  10), 

(x  —  a)(x-b)f'(x)       .       ,  v       .        N  .  .        w       MQ'0*0 
f(x)  -=r(g-b)+8(a;-o)  +  (a;-o)(a;-6)Q^. 

The  second  member  has  the  value  r(a  —  b)<0  for  x  =  a  and  the  value 
s(b— a)>0  for  re  =  6,  and  hence  vanishes  an  odd  number  of  times  between 
a  and  b  (§63).  But,  in  the  left  member,  (x—a)  (x  —  b)  and/(:r)  remain 
of  constant  sign  between  a  and  b,  since  f(x)  =  0  has  no  root  between  a 
and  6.     Hence  f'(x)  vanishes  an  odd  number  of  times. 

Corollary.  Between  two  consective  real  roots  a  and  /3  of  f'(x)=0 
there  occurs  at  most  one  real  root  of  f(x)  =0. 

For,  if  there  were  two  such  real  roots  a  and  b  of  f(x)  =  0,  the  theorem 
shows  that  f'(x)  =0  would  have  a  real  root  between  a  and  b  and  hence  be- 
tween a  and  /3,  contrary  to  hypothesis. 

Applying  also  §  63  we  obtain  the 

Criterion.  If  a  and  /3  are  consecutive  real  roots  of  f'(x)  =  0,  then  f(x)  =  0 
has  a  single  real  root  between  a  and  /3  if  f(a)  and  f(/3)  have  opposite  signs, 
but  no  root  if  they  have  like  signs.  At  most  one  real  root  of  f(x)  =0  is  greater 
than  the  greatest  real  root  of  f'(x)  =0,  and  at  most  one  real  root  of  f(x)  =0  is 
less  than  the  least  real  root  of  f'(x)  =0. 


70  GRAPHS  [Ch.  V 

If  f(a)=0  for  our  root  a  of  f{x)=Q,  a  is  a  multiple  root  of  f(x)=0 
and  it  would  be  removed  before  the  criterion  is  applied. 

Example.     For/(x)  =  3x5-25x3+60x-20, 

IL/'(a;)=X4_5x2+4  =  (a.2_1)    (£2_4)t 

Hence  the  roots  of  f'(x)  =0  are  ±1,  ±2.     Now 

f(_oo)  =  -oo,    /(-2)  =  -36,    /(-l)  =  -58,    /(1)=18,    /(2)  =  -4,    /(+«))  =  +  oo. 

Hence  there  is  a  single  real  root  in  each  of  the  intervals 

.     (-1,   1),         (1,  2),         (2,  + co), 
and  two  imaginary  roots.     The  three  real  roots  are  positive. 

EXERCISES 

1.  Prove  that  x5  — 5x+2=0  has  1  negative,  2  positive  and  2  imaginary  roots. 

2.  Prove  that  x6+x  — 1=0  has  1  negative,  1  positive  and  4  imaginary  roots. 

3.  Show  that  x5  —  3x3+2x2  —  5  =  0  has  two  imaginary  roots,  and  a  real  root  in  each 
of  the  intervals  (-2,  -1.5),  (-1.5,  -1),  (1,  2). 

4.  Prove  that  4x5— 3x4— 2x2+4x  — 10  =  0  has  a  single  real  root. 

5.  Show  that,  if  /(S)(x)  =0  has  imaginary  roots,  /(x)  =0  has  imaginary  roots. 

6.  Derive  Rolle's  theorem  from  the  fact  that  there  is  an  odd  number  of  bend  points 
between  a  and  b,  the  abscissa  of  each  being  a  root  of /'(x)  =0  of  odd  multiplicity,  while 
the  abscissa  of  an  inflexion  point  with  a  horizontal  tangent  is  a  root  of /'(x)  =0  of  even 
multiplicity, 


CHAPTER  VI 
Isolation  of  the  Real  Roots  of  a  Real  Equation 

66.  Purpose  and  Methods  of  Isolating  the  Real  Roots.  In  the  next 
chapter  we  shall  explain  processes  of  computing  the  real  roots  of  a  given 
real  equation  to  any  assigned  number  of  decimal  places.  Each  such 
method  requires  some  preliminary  information  concerning  the  root  to 
be  computed.  For  example,  it  would  be  sufficient  to  know  that  the  root 
is  between  4  and  5,  provided  there  be  no  other  root  between  the  same 
limits.  But  in  the  contrary  case,  narrower  limits  are  necessary,  such 
as  4  and  4.3,  with  the  further  fact  that  only  one  root  is  between  these  new 
limits.     Then  that  root  is  said  to  be  isolated. 

If  an  equation  has  a  single  positive  root  and  a  single  negative  root,  the  real  roots 
are  isolated,  since  there  is  a  single  root  between  —  oo  and  0,  and  a  single  one  between 
0  and  +  oo  .  However,  for  the  practical  purpose  of  their  computation,  we  shall  need 
narrower  limits,  sufficient  to  fix  the  first  significant  figure  of  each  root,  for  example 
-40  and  -30,  or  20  and  30. 

We  may  isolate  the  real  roots  of  f(x)  =  0  by  means  of  the  graph  of 
y=f(x).  But  to  obtain  a  reliable  graph,  we  saw  in  Chapter  V  that  we 
must  employ  the  bend  points,  whose  abscissas  occur  among  the  roots 
f(x)=0.  Since  the  latter  equation  is  of  degree  n—l  when  f(x)=0  is  of 
degree  n,  this  method  is  usually  impracticable  when  n  exceeds  3.  The 
method  based  on  Rolle's  theorem  (§  65)  is  open  to  the  same  objection. 

The  most  effective  method  is  that  due  to  Sturm  (§68).  We  shall, 
however,  begin  with  Descartes'  rule  of  signs  since  it  is  so  easily  applied. 
Unfortunately  it  rarely  tells  us  the  exact  number  of  real  roots. 

67.  Descartes'  Rule  of  Signs.  Two  consecutive  terms  of  a  real  poly- 
nomial or  equation  are  said  to  present  a  variation  of  sign  if  their  coefficients 
have  unlike  signs.  By  the  variations  of  sign  of  a  real  polynomial  or  equa- 
tion we  mean  all  the  variations  presented  by  consecutive  terms. 

Thus,  in  x5—  2x3  — 4x2+3  =  0,  the  first  two  terms  present  a  variation  of  sign,  and 
likewise  the  last  two  terms.     The  number  of  variations  of  sign  of  the  equation  is  two. 

71 


72  ISOLATION  OF  REAL  ROOTS  [Ch.  VI 

Descartes'  Rule.  The  number  of  positive  real  roots  of  an  equation 
with  real  coefficients  is  either  equal  to  the  number  of  its  variations  of  sign 
or  is  less  than  that  number  by  a  positive  even  integer.  A  root  of  multiplicity 
m  is  here  counted  as  m  roots. 

For  example,  xG—Sx2+x-\-l  =0  has  either  twoior  no  positive  roots,  the  exact  number 
not  being  found.  But  3a;3—  x  — 1=0  has  exactly  one  positive  root,  which  is  a  simple 
root. 

Descartes'  rule  will  be  derived  in  §  73  as  a  corollary  to  Budan's  theorem. 
The  following  elementary  proof  1  was  communicated  to  the  author  by 
Professor  D.  R.  Curtiss. 

Consider  any  real  polynomial 

f(x)  =  a0xn+a1xn-1+  .  .  .  +alxn~%  (oo^O,  a,3*0). 

Let  r  be  a  positive  real  number.     By  actual  multiplication, 

F{x)^{x-r)f{x)=AQxn+l+Aixn+  .  .  .  +Al+lxn~l, 
where 

Ao  =  ao,  Ai=ai—rao,       A2  =  a2  —  rai,  .  .  .  ,  Al  =  ai—ral^l,  Al+1  =  —ral. 

In  f(x)  let  atl  be  the  first  non-vanishing  coefficient  of  different  sign  from 
oq,  let  ai2  be  the  first  non-vanishing  coefficient  following  atl  and  of  the 
same  sign  as  ao,  etc.,  the  last  such  term,  akv,  being  either  at  or  of  the  same 
sign  as  a%.     Evidently  v  is  the  number  of  variations  of  sign  of  fix). 

For  example,  if/(x)  =2x6+3x5—  4x4— 6x3-r-7x,  we  have  y  =  2,  at,x=a%=*  —4,  at2=ah =7. 
Note  that  04  =  0  since  x2  is  absent. 

The  numbers  Ao,  Atv  .  .  .  ,  A^,  Al+i  are  all  different  from  zero  and 
have  the  same  signs  as  ao,  a4l,  .  .  .  ,  atv,  —ah  respectively.  This  is 
obviously  true  for  Aq  =  oo  and  Ai+i=  —rah  Next,  A^  is  the  sum  of  the 
non- vanishing  number  ati  and  the  number  —  ra^x,  which  is  either  zero 
or  else  of  the  same  sign  as  ati  since  a^-i  is  either  zero  or  of  opposite  sign 
to  ati.    Hence  the  sum  Ati  is  not  zero  and  has  the  same  sign  as  aki. 

By  hypothesis,  each  of  the  numbers  ao,  atv  .  .  .  ,  akv  after  the  first 
is  of  opposite  sign  to  its  predecessor,  while  —  a}  is  of  opposite  sign  to  at(). 
Hence  each  term  after  the  first  in  the  sequence  Ao,  Atl,  .  .  .  ,  Atv,  Al+1 
is  of  opposite  sign  to  its  predecessor.  Thus  these  terms  present  v-\-l 
variations  of  sign.  We  conclude  that  F(x)  has  at  least  one  more  vari- 
ation of  sign  than  f{x).     But  we  may  go  further  and  prove  the  following 

1  The  proofs  given  in  college  algebras  are  mere  verifications  of  special  cases. 


§  67]  DESCARTES'  RULE  OF  SIGNS  73 

Lemma.  The  number  of  variations  of  sign  of  F(x)  is  equal  to  that  of 
f(x)  increased  by  some  positive  odd  integer. 

For,  the  sequence  Aq,  A\,  .  .  .  ,  Akl  has  an  odd  number  of  variations 
of  sign  since  its  first  and  last  terms  are  of  opposite  sign;    and  similarly 

for  the  v  sequences 

Aiv      Atl+i,  .  .  .  ,  Ai2; 


Aiv,      Atv+1,  .  .  .  ,  Al+1. 

The  total  number  of  variations  of  sign  of  the  entire  sequence  Aq,  A\,  .  .  .  , 
Al+1  is  evidently  the  sum  of  the  numbers  of  variations  of  sign  for  the 
y+1  partial  sequences  indicated  above,  and  is  thus  the  sum  of  v+1  posi- 
tive odd  integers.  Since  each  such  odd  integer  may  be  expressed  as  1 
plus  0  or  a  positive  even  integer,  the  sum  mentioned  is  equal  to  v+1  plus 
0  or  a  positive  even  integer,  i.e.,  to  v  plus  a  positive  odd  integer. 

To  prove  Descartes'  rule  of  signs,  consider  first  the  case  in  which  f(x)  =  0 
has  no  positive  real  roots,  i.e.,  no  real  root  between  0  and  +oo.  Then 
/(0)  and  /(oo)  are  of  the  same  sign  (§  63),  and  hence  the  first  and  last 
coefficients  of  f(x)  are  of  the  same  sign.1  Thus  f{x)  has  either  no  vari- 
ations of  sign  or  an  even  number  of  them,  as  Descartes'  rule  requires. 

Next,  let  f{x)  =  0  have  the  positive  real  roots  n,  .  .  .  , rfc  and  no  others. 
A  root  of  multiplicity  m  occurs  here  m  times,  so  that  the  r's  need  not  be 
distinct.     Then 

f(x)  =  (x-n)  .  .  .  (x-rk)(p(x), 

where  <f>(x)  is  a  polynomial  with  real  coefficients  such  that  4>(x)=0  has 
no  positive  real  roots.  We  saw  in  the  preceding  paragraph  that  4>(x) 
has  either  no  variations  of  sign  or  an  even  number  of  them.  By  the 
Lemma,  the  product  (x  —  rk)4>(x)  has  as  the  number  of  its  variations  of 
sign  the  number  for  <f>(x)  increased  by  a  positive  odd  integer.  Similarly 
when  we  introduce  each  new  factor  x  —  rt.  Hence  the  number  of  varia- 
tions of  sign  of  the  final  product  f(x)  is  equal  to  that  of  </>(a;)  increased 
by  k  positive  odd  integers,  i.e.,  by  k  plus  0  or  a  positive  even  integer. 
Since  <j>(x)  has  either  no  variations  of  sign  or  an  even  number  of  them, 
the  number  of  variations  of  sign  of  f(x)  is  k  plus  0  or  a  positive  even  integer, 
a  result  equivalent  to  our  statement  of  Descartes'  rule. 

1  In  case  f(x)  has  a  factor  xn~l,  we  use  the  polynomial  f(x)/xn~  instead  of  f(x)  in 
this  argument. 


74  ISOLATION  OF  REAL  ROOTS  [Ch.  VI 

If  —  p  is  a  negative  root  of  f(x)  =  0,  then  p  is  a  positive  root  of  /(  —  x)  =  0. 
Hence  we  obtain  the 

Corollary.  The  number  of  negative  roots  of  f(x)  =  0  is  either  equal 
to  the  number  of  variations  of  sign  of  f(—x)  or  is  less  than  that  number 
by  a  positive  even  integer. 

For  example,  x4+3x3+x  — 1=0  has  a  single  negative  root,  which  is  a  simple  root, 
since  x4  —  3x3  —  x  —  1  =0  has  a  single  positive  root. 

As  indicated  in  Exs.  10,  11  below,  Descartes'  rule  may  be  used  to  isolate 
the  roots. 

EXERCISES 

Prove  by  Descartes'  rule  the  statements  in  Exs.  1-8,  12,  15. 

1.  An  equation  all  of  whose  coefficients  are  of  like  sign  has  no  positive  root.  Why 
is  this  self-evident? 

2.  There  is  no  negative  root  of  an  equation,  like  x5  — 2x4  —  3x2+7x  —  5  =  0,  in  which 
the  coefficients  of  the  odd  powers  of  x  are  of  like  sign,  and  the  coefficients  of  the  even 
powers  (including  the  constant  term)  are  of  the  opposite  sign.  Verify  by  taking  x  =  —  p, 
where  p  is  positive. 

3.  x3+a2x+i>2  =  0  has  two  imaginary  roots  if  b^O. 

4.  For  n  even,  xn  —  1  =0  has  only  two  real  roots. 

5.  For  n  odd,  xn  —  1  =0  has  only  one  real  root. 

6.  For  n  even,  xn+l  =0  has  no  real  root;  for  n  odd,  only  one. 

7.  x4+12x2+5x  — 9  =  0  has  just  two  imaginary  roots. 

8.  xi-\-a2x2+b2x  — c2  =  0  (c^O)  has  just  two  imaginary  roots. 

9.  Descartes'  rule  enables  us  to  find  the  exact  number  of  positive  roots  only  when 
all  the  coefficients  are  of  like  sign  or  when 

f(x)=XU  +  piXn~1+   .   .   .   +pn-sXS  —  pn-s  +  iXS~1-    .   .   .    —pn  =  0, 

each  pi  being  ^0.  Without  using  that  rule,  show  that  the  latter  equation  has  one 
and  only  one  positive  root  r.  Hints:  There  is  a  positive  root  r  by  §  63  (a  =  0,  b=  oo  ). 
Denote  by  P(x)  the  quotient  of  the  sum  of  the  positive  terms  by  xs,  and  by  —N(x) 
that  of  the  negative  terms.  Then  N(x)  is  a  sum  of  powers  of  1/x  with  positive  coef- 
ficients. 

If     x>r,         P(x)>P(r),         N(x)<N(r),        /(x)>0; 
If     x<r,         P(x)<P(r),         N(x)>N(r),         /(x)<0. 

10.  Prove  that  we  obtain  an  upper  limit  to  the  number  of  real  roots  of  /(x)=0 
between  a  and  b,  if  we  set 

a-\-by  /  x  —  a 

'■y 


l+y  \  b—xj 

multiply  by  (l+y)n,  and  apply  Descartes'  rule  to  the  resulting  equation  in  y. 


§  68]  STURM'S  METHOD  75 

11.  Show  by  the  method  of  Ex.  10  that  there  is  a  single  root  between  2  and  4  of 
x3+z2-17x+15  =  0.     Here  we  have  27y3+Zy2-23y-7  =  0. 

12.  In  the  astronomical  problem  of  three  bodies  occurs  the  equation 

r5  +  (3-M)r4+(3-2M)r3-Mr2-2Mr-M  =  0, 

where  0</x<l.     Why  is  there  a  single  positive  real  root? 

13.  Prove  that  xb+x3  —  x2+2x— 3  =  0  has  four  imaginary  roots  by  applying  Des- 
cartes' rule  to  the  equation  in  y  whose  roots  are  the  squares  of  the  roots  of  the  former. 
Transpose  the  odd  powers,  square  each  new  member,  and  replace  x2  by  y. 

14.  As  in  Ex.  13  prove  that  x3+x2+8x+6  =  0  has  imaginary  roots. 

15.  If  a  real  equation  f(x)  =0  of  degree  n  has  n  real  roots,  the  number  of  positive 
roots  is  exactly  equal  to  the  number   V  of  variations  of  sign.     Hint:    consider  also 

16.  Show  that  x3—  x2+2x  +  l=0  has  no  positive  root.     Hint:  multiply  by  x+1. 

68.  Sturm's  Method.  Let  f(x)  =  0  be  an  equation  with  real  coefficients, 
and  f{x)  the  first  derivative  of  f{x).  The  first  step  of  the  usual  process 
of  finding  the  greatest  common  divisor  of  f(x)  and  fix),  if  it  exists,  con- 
sists in  dividing  /  by  /'  until  we  obtain  a  remainder  r(x),  whose  degree 
is  less  than  that  of/'.  Then,  if  q\  is  the  quotient,  we  have  f=qif-\-r. 
Instead  of.  dividing  /'  by  r,  as  in  the  greatest  common  divisor  process,  and 
proceeding  further  in  that  manner,  we  write  fa=—r,  divide  /'  by  fa,  and 
denote  by  fa  the  remainder  with  its  sign  changed.     Thus 

f=Qif-fa,      /'= 22/2-/3,      fa  =  93/3-/4, 

The  latter  equations,  in  which  each  remainder  is  exhibited  as  the  nega- 
tive of  a  polynomial  fa  yield  a  modified  process,  just  as  effective  as  the 
usual  process,  of  finding  the  greatest  common  divisor  G  of  f(x)  and /'(a;)  if 
it  exists. 

Suppose  that  —fa  is  the  first  constant  remainder.  If  fa  =  0,  then  /3  =  G, 
since  fa  divides  fa  and  hence  also  f  and  /  (as  shown  by  using  our  above 
equations  in  reverse  order);  while,  conversely,  any  common  divisor  of 
/  and  /'  divides  fa  and  hence  also  /3. 

But  if  fa  is  a  constant  5^0,  /  and/'  have  no  common  divisor  involving 
x.  This  case  arises  if  and  only  if  f(x)=0  has  no  multiple  root  (§  58), 
and  is  the  only  case  considered  in  §§  69-71. 

Before  stating  Sturm's  theorem  in  general,  we  shall  state  it  for  a 
numerical  case  and  illustrate  its  use, 


76 


ISOLATION  OF  REAL  ROOTS 


[Cb.  VI 


Example.    f(x)  =  z3+4z2-7.     Then/'  =  3:c2+8z, 


/=(^+i)/'-/2, 


f2  =  ™x+7, 


_  422  1 
1  0  2  4  • 


For  1  x  =  l,  the  signs  of  /,  /',  /2,  /3,  are h  +  +,  showing  a  single  variation  of 

consecutive  signs.     For  x  =  2,  the  signs  are  -\ — | — | — \-,  showing  no  variation  of  sign. 
Sturm's  theorem  states  that  there  is  a  single  real  root  between  1  and  2.     For  x=  —  oo  , 

the  signs  are  —  -\ \-,  showing  3  variations  of  sign.     The  theorem  states  that  there 

are  3  —  1=2  real  roots  between  —  oo  and  1 .     Similarly, 


X 

Signs 

Variations 

- 1 

-  2 

-  3 
-4 

1    ++    1 
+  +    1     1 

111    + 
+  +  +  + 

1 

2 
2 
3 

Hence  there  is  a  single  real  root  between  —2  and  —1,  and  a  single  one  between  —4 
and  —3.  Each  real  root  has  now  been  isolated  since  we  have  found  two  numbers 
such  that  a  single  real  root  lies  between  these  two  numbers  or  is  equal  to  one  of  them. 

Some  of  the  preceding  computation  was  unnecessary.  After  isolating  a  root  between 
—2  and  —1,  we  know  that  the  remaining  root  is  isolated  between  —  oo  and  —2.  But 
before  we  can  compute  it  by  Horner's  method,  we  need  closer  limits  for  it.  For  that 
purpose  it  is  unnecessary  to  find  the  signs  of  all  four  functions,  but  merely  the  sign 
of/ (§63). 

69.  Sturm's  Theorem.  Let  f{x)  =  0  be  an  equation  with  real  coefficients 
and  without  multiple  roots.  Modify  the  usual  process  of  seeking  the  great- 
est common  divisor  of  f(x)  and  its  first  derivative  2  f\  (x)  by  exhibiting  each 
remainder  as  the  negative  of  a  polynomial  /4; 

(1)      f=Qlfl—f2,        fl  =  q2f2—f3,       f2  =  qsh~ /*,  •  •  •  ,      fn-2=qn-lfn-l-fn, 

where 3  /„  is  a  constant ^0.     If  a  and  b  are  real   numbers,    a<b,   neither 

1  Before  going  further,  check  that  the  preceding  relations  hold  when  a>=  1  by  insert- 
ing the  computed  values  of  /,  /',  /*  for  x  =  1 .  Experience  shows  that  most  students  make 
some  error  in  finding  /2,  /3,  .  .  .  ,  so  that  checking  is  essential. 

2  The  notation  /i  instead  of  the  usual  /',  and  similarly  /o  instead  of  /,  is  used  to  reg- 
ularize the  notation  of  all  the  f's,  and  enables  us  to  write  any  one  of  the  equations  (1) 
in  the  single  notation  (3) . 

3  If  the  division  process  did  not  yield  ultimately  a  constant  remainder  5^0,  /  and  /» 
would  have  a  common  factor  involving  x,  and  hence  f(x)  =0  a  multiple  root. 


§*69]  STURM'S  THEOREM  77 

a  root  of  f{x)  =  0,  the  number  of  real  roots  of  f{x)  =0  between  a  and  b  is  equal 
to  the  excess  of  the  number  of  variations  of  sign  of 

(2)  /Or),        fx(x),        f2(x),     ...-,    /,_!(*),        /. 

for  x  =  a  over  the  number  of  variations  of  sign  for  x  =  b.  Terms  which  vanish 
are  to  be  dropped  out  before  counting  the  variations  of  sign. 

For  brevity,  let  Vx  denote  the  number  of  variations  of  sign  of  the 
numbers  (2)  when  x  is  a  particular  real  number  not  a  root  of  f(x)  =  0. 

First,  if  x\  and  X2  are  real  numbers  such  that  no  one  of  the  continuous 
functions  (2)  vanishes  for  a  va.iue  of  x  between  x\  and  X2  or  for  x  =  x\  or 
x  =  x%,  the  values  of  any  one  of  these  functions  for  x  =  xi  and  x  =  X2  are 
both  positive  or  both  negative  (§  63).  and  therefore  VXl=VX2. 

Second,  let  p  be  a  root  of  ft(x)  =  0,  where  l£i<n.     Then 

(3)  ft-  iO)  =  Qifi(x)  -fi+J(x) 

and  the  equations  (1)  following  this  one  show  that  ft-i(x)  and  ft(x)  have 
no  common  divisor  involving  x  (since  it  would  divide  the  constant  /„). 
By  hypothesis,  ft(x)  has  the  factor  x  —  p.  Hence  ft-iix)  does  Dot  have 
this  factor  x—p.     Thus,  by  (3), 

/«-i(p)  =  -/t+i(p)^0. 

Hence,  if  p  is  a  sufficiently  small  positive  number,  the  values  of 

/«- iG»Di        ft(x),        ft+i(x) 

for  x=p  —  p  show  just  one  variation  of  sign,  since  the  first  and  third 
values  are  of  opposite  sign,  and  for  x  =  p-\-p  show  just  one  variation  of 
sign,  and  therefore  show  no  change  in  the  number  of  variations  of  sign 
for  the  two  values  of  x. 

It  follows  from  the  first  and  second  cases  that  Va  =  Vp  if  a  and  /3  are 
real  numbers  for  neither  of  which  any  one  of  the  functions  (2)  vanishes 
and  such  that  no  root  of  f(x)  =  0  lies  between  a  and  /3. 

Third,  let  r  be  a  root  of  fix)  =  0.     By  Taylor's  theorem  (8)  of  §  56, 

/(r-p)  =  -#(r)+M"(r)-  ..., 
f(r+p)=     vf(r)+hv2f"(S)+  .  .  .    . 

If  p  is  a  sufficiently  small  positive  number,  each  of  these  polynomials  in 
p  has  the  same  sign  as  its  first  term.     For,  after  removing  the  factor  p, 


78  ISOLATION  OF  REAL  ROOTS  [Ch.  VI 

we  obtain  a  quotient  of  the  form  ao+s,  where  s  =  aip+«2f2+  ...  is 
numerically  less  than  ao  for  all  values  of  p  sufficiently  small  (§  62).  Hence 
if  f'(r)  is  positive,  f(r  —  p)  is  negative  and  /(r+p)  is  positive,  so  that  the 

terms  j(x),  fi(x)=f(x)  have  the  signs (-  for  x  =  r  —  p  and  the  signs 

H — h  for  x  —  r-\-p.     If  /'(r)  is  negative,  these  signs  are  -| and  —  — 

respectively.  In  each  case,  f(x),  f\(x)  show  one  more  variation  of  sign 
for  x  =  r  —  p  than  for  x  =  r-\-p.  Evidently  p  may  be  chosen  so  small  that 
no  one  of  the  functions /i(x),  .  .  .  ,fn  vanishes  for  either  x  =  r  —  p  or  x  =  r-\-p, 
and  such  that  fi(x)  does  not  vanish  for  a  value  of  x  between  r  —  p  and 
r-\-p,  so  that  f(x)=0  has  the  single  real  root  r  between  these  limits  (§  65). 
Hence  by  the  first  and  second  cases,  /i,  ...,/„  show  the  same  number 
of  variations  of  sign  for  x  =  r  —  p  as  for  x  =  r-\-p.  Thus,  for  the  entire 
series  of  functions  (2),  we  have 

(4)  VT-P-Vr+P=l. 

The  real  roots  of  f(x)  =  0  within  the  main  interval  from  a  to  b  (i.e.,  the 
aggregate  of  numbers  between  a  and  b)  separate  it  into  intervals.  By 
the  earlier  result,  Vx  has  the  same  value  for  all  numbers  in  the  same 
interval.  By  the  present  result  (4),  the  value  of  Vx  in  any  interval  exceeds 
the  value  for  the  next  interval  by  unity.  Hence  Va  exceeds  Vb  by  the 
number  of  real  roots  between  a  and  b. 

Cokollary.     If  a<b,  then  Va=  Vb. 

A  violation  of  this  Corollary  usually  indicates  an  error  in  the  com- 
putation of  Sturm's  functions  (2). 

EXERCISES 

Isolate  by  Sturm's  theorem  the  real  roots  of 

1.  x3+2z+20  =  0.  2.  .t3+x-3  =  0. 

70.  Simplifications  of  Sturm's  Functions.  In  order  to  avoid  fractions, 
we  may  first  multiply  f(x)  by  a  positive  constant  before  dividing  it  by 
fi(x),  and  similarly  multiply  /i  by  a  positive  constant  before  dividing  it 
by  /2,  etc.  Moreover,  we  may  remove  from  any  /<  any  factor  kt  which  is 
either  a  positive  constant  or  a  polynomial  in  x  positive  for  1  a^LxS. b, 
and  use  the  remaining  factor  Ft  as  the  next  divisor. 

To  prove  that  Sturm's  theorem  remains  true  when  these  modified 

1  Usually  we  would  require  that  hi  be  positive  for  all  values  of  x,  since  we  usually 
wish  to  employ  the  limits  —  oo  and  +  oo  . 


§  70]  SIMPLIFICATIONS  OF  STURM'S  FUNCTIONS  79 

functions/,  F\,  .  .  .  ,  Fm  are  employed  in  place  of  functions  (2),  consider 
the  equations  replacing  (1): 

fi  =  faF1}         c2f=qiFi-k2F2,         c3Fi  =  q2F2-k3F3, 

c±F2  =  qzF3  —  kiF±,  .  .  .  ,  cmFm_2  =  qm-iFm_1  —  kmFm, 

in  which  c2,  c3,  .  .  .  are  positive  constants  and  Fm  is  a  constant  5^0.  A 
common  divisor  (involving  x)  of  Ft_x  and  Ft  would  divide  Fi_2,  .  .  .  , 
F2,  F\,  f,  fi,  whereas  f(x)=0  has  no  multiple  roots.  Hence  if  p  is  a  root 
of  Ft(x)  =  0,  then  F^p) ^0  and 

A+iFi-i(p)--h+i(f>)Fi+i(p),        ci+1>0,        ki+1(P)>0. 
Thus  Ft_i  and  Fi+l  have  opposite  signs  for  x  =  p.     We  proceed  as  in  §  69. 

Example  1.  If  f{x)  =x3+6x  — 10,  /i  =  3(a;2+2)  is  always  positive.  Hence  we  may 
employ  /  and  Fi  =  1 .  For  x  =  —  <x> ,  there  is  one  variation  of  sign ;  for  x  =  +  oo,  no 
variation.     Hence  there  is  a  single  real  root;   it  lies  between  1  and  2. 

Example  2.     If  f(x)  =  2xi  —  13x2  —  lOx  — 19,  we  may  take 

/i=4a;3-13x-5. 
Then 

2/=x/1-/2,        /2  =  13x2  +  15x+38  =  13(x+if)2+iffi 

Since /2  is  always  positive,  we  need  go  no  further  (we  may  take  F2  —  l).  For  x=  —  oo , 

the  signs  are  -\ 1-;    forx  =  +  oo,+  +  +.     Hence  there  are  two  real  roots.     The 

signs  for  x  =  0  are  — h     Hence  one  real  root  is  positive  and  the  other  negative. 

EXERCISES 

Isolate  by  Sturm's  theorem  the  real  roots  of 

1.  x3+3x2-2x-5  =  0.  2.  x4  +  12x2+5x-9  =  0. 

3.  x3-7x-7=0.  4.  3x4-6x2+8x-3=0. 

5.  x6+6x5-30x2-12x-9=0  [stop  with  /2] . 

6.  x4-8x3+25x2-36x+8  =  0. 

7.  For/  =  x3+px+g  (p^0),  show  that/i  =  3x2+p,         f2=  -2px-3g, 

4p2/i  =  ( -  Gpx  +9g)/,  -fa  /3  =  -  4p3  -  27g2, 

so  that  fz  is  the  discriminant  A  (§  44).  Let  [p]  denote  the  sign  of  p.  Then  the  signs 
of  /,  /i,  ft,  /s  are 

-      +      +[p]     [A]         forx=-oo, 

+      +      -[p]     [A]         forx=  +  ^o. 

For  A  negative  there  is  a  single  real  root.  For  A  positive  and  therefore  p  negative, 
there  are  three  distinct  real  roots.  For  A  =  0,  f%  is  a  divisor  of /i  and/,  so  that  x  = 
—  3q/  (2p)  is  a  double  root. 


80  ISOLATION  OF  REAL  ROOTS  [Ch.  VI 

8.  Prove  that  if  one  of  Sturm's  functions  has  p  imaginary  roots,  the  initial  equation 
has  at  least  p  imaginary  roots. 

9.  State  Sturm's  theorem  so  as  to  include  the  possibility  of  a,  or  b,  or  both  a  and  b 
being  roots  of  f(x)  =0. 

71.  Sturm's  Functions  for  a  Quartic  Equation.    For  the  reduced  quar- 
tic  equation  f(z)  =  0, 

'  /  =  z4L-{-qz2-\-rz-\-s, 

(5)  \  fi=4z*+2qz+r, 

.  J2=  —  2qz2  —  3rz— 4s. 
Let  q9^0  and  divide  q2fi  by/2.     The  negative  of  the  remainder  is 

(6)  f3  =  Lz-12rs-rq2,         L  =  Sqs-2q3-9r2. 

Let  L=^0.  Then  f±  is  a  constant  which  is  zero  if  and  only  if  /=0  has 
multiple  roots,  i.e.,  if  its  discriminant  A  is  zero.  We  therefore  desire  /t 
expressed  as  a  multiple  of  A.     By  §  50, 

(7)  A=-4P3-27Q2,        P=-4s-^,        Q  =  ^qs-r2-^. 
We  may  employ  P  and  Q  to  eliminate 

(8)  4s=-P-|2,         r2=-Q-%qP-^Tqs. 

We  divide  L2J2  by 

(9)  f3  =  Lz+3rP,        L  =  9Q+4gP. 

The  negative  of  the  remainder x  is 

(10)  18r2qP2-  9r2LP+4sL2  =  q2A. 

The  left  member  is  easily  reduced  to  q2A.  Inserting  the  values  (8)  and 
replacing  L2  by  L(9Q+4gP),  we  get 

-  18gQP2  -  12q2P3  -  i£q4P2+2qP2L+±q3PL  -  3q2QL. 

Replacing  L  by  its  value  (9),  we  get  q2A.     Hence  we  may  take 

(11)  /4  =  A. 

Hence  if  gLA^O,  we  may  take  (5),  (9),  (11)  as  Sturm's  functions. 

1  Found  directly  by  the  Remainder  Theorem  (§  14)  by  inserting  the  root  2=  —SrP/L 
of /3  =  0  into  L2/2. 


§  71]  STURM'S  FUNCTIONS  FOR  A  QUARTIC  81 

Denote  the  sign  of  q  by  [q\.     The  signs  of  Sturm's  functions  are 
+     -      -[q]       -[L]     [A]     forx=-  oo, 

+     +     -k]  [L]     [A]     iorx=+oo. 

First,  let  A>0.     If  q  is  negative  and  L  is  positive,  the  signs  are 

H 1 h  and  +  +  +  +  +  ,  so  that  there  are  four  real  roots.     In  each 

of  the  remaining  three  cases  for  q  and  L,  there  are  two  variations  of  sign 
in  either  of  the  two  series  and  hence  there  is  no  real  root. 

Next,  let  A<0.  In  each  of  the  three  cases  in  which  q  and  L  are  not 
both  positive,  there  are  three  variations  of  sign  in  the  first  series  and  one 
variation  in  the  second,  and  hence  just  two  real  roots.  If  q  and  L  are 
both  positive,  the  number  of  variations  is  1  in  the  first  series  and  3  in  the 
second,  so  that  this  case  is  excluded  by  the  Corollary  to  Sturm's  theorem. 
To  give  a  direct  proof,  note  that,  by  the  value  of  L  in  (6),  L>0,  q>0 
imply  4s>32,  i.e.,  s>0,  and  hence,  by  (7),  P  is  negative,  so  that  each  term 
of  (10)  is  2:0,  whenceA>0. 

Hence,  if  gLA^O,  there  are  four  distinct  real  roots  if  and  only  if  A 
and  L  are  positive,  and  q  negative;  two  distinct  real  and  two  imaginary 
roots  if  and  only  if  A  is  negative. 

Combining  this  result  with  that  in  Ex.  4  below,  we  obtain  the 

Theorem.  If  the  discriminant  A  of  zt+qz2-\-rz-\-s  =  0  is  negative,  there 
are  two  distinct  real  roots  and  two  imaginary  roots;  if  A>0,  q<0,  L>0, 
four  distinct  real  roots;  if  A > 0  and  either  qhO  or  L £ 0,  no  real  roots.  Here 
L  =  8qs-2q3-9r2. 

Our  discussion  furnished  also  the  series  of  Sturm  functions,  which 
may  be  used  in  isolating  the  roots. 

EXERCISES 

1.  If  qA^O,  L  =  0,  then  /3  =  3rP  is  not  zero  (there  being  no  multiple  root)  and  its 
sign  is  immaterial  in  determining  the  number  of  real  roots.  Prove  that  there  are  just 
two  real  roots  if  q<0,  and  none  if  q>0.     By  (10),  q  has  the  same  sign  as  A. 

2.  If  r A  5^0,  q  =  0,  obtain  — f3  by  substituting  z  =  —  4s/(3r)  in  /i.  Show  that  we  may 
take/3  =  rA  and  that  there  are  just  two  real  roots  if  A<0,  and  no  real  roots  if  A>0. 

3.  If  A=^0,  </  =  r  =  0,  prove  that  there  are  just  two  real  roots  if  A<0,  and  no  real 
roots  if  A>0.     Since  A=256s3,  check  by  solving  z4-|-.s  =  0. 

4.  If  A  7^0,  qL  =  0,  there  are  just  two  real  roots  if  A<0,  and  no  real  roots  if  A>0. 
[Combine  the  results  in  Exs.  1-3.] 

5.  Apply  the  theorem  to  Exs.  2,  4,  6  of  §  70. 

6.  Isolate  the  real  roots  of  Exs.  3,  4,  5  of  §  48. 


82  ISOLATION  OF  REAL  ROOTS  [Ch.  VI 

72.  Sturm's  Theorem  for  the  Case  of  Multiple  Roots.  We  might 
remove  the  multiple  roots  by  dividing /(z)  by1  fn(x),  the  greatest  com- 
mon divisor  of  fix)  and  f\  =f'(x) ;  but  this  would  involve  considerable 
work,  besides  wasting  the  valuable  information  in  hand.  As  before,  we 
suppose /(a)  and  /(&)  different  from  zero.  We  have  equations  (1)  in 
which  /„  is  now  not  a  constant. 

The  difference  Va—Vb  is  the  number  of  real  roots  between  a  and  b,  each 
multiple  root  being  counted  only  once.      (i[^ 

If  p  is  a  root  of ft(x)  =  0,  but  not  a  multiple  root  of  f(x)—0,  then /<_i(p) 
is  not  zero.  For,  if  it  were  zero,  x  —  p  would  by  (1)  be  a  common  factor 
of  /  and  /i.     We  may  now  proceed  as  in  the  second  case  in  §  69. 

The  third  case  requires  a  modified  proof  only  when  r  is  a  multiple  root. 
Let  r  be  a  root  of  multiplicity  m,  ra^.2.  Then./(r),  f(r),  .  .  .  ,  f{m~l)(r) 
are  zero  and,  by  Taylor's  theorem, 

f'(r-\-v)  = f(m)0)  + 

J  v^W     1-2.  .  .  (m-iy     K)^ 

These  have  like  signs  if  p  is  a  positive  number  so  small  that  the  signs  of 
the  polynomials  are  those  of  their  first  terms.  Similarly,  f(r  —  p)  and 
f(r—p)  have  opposite  signs.  Hence  /  and  f\  show  one  more  variation 
of  sign  for  x  =  r  —  p  than  for  x^r+p.  Now  (x— r)w_1  is  a  factor  of  / 
and  /i  and  hence,  by  (1),  of  f2,  .  .  .  ,  /„.  Let  their  quotients  by  this 
factor  be  <t>,  <£i,  .  .  •  ,  4>n-  Then  equations  (1)  hold  after  the/'s  are  replaced 
by  the  4>'s.  Taking  p  so  small  that  4>iix)=0  has  no  root  between  r  —  p 
and  r-\-p,  we  see  by  the  first  and  second  cases  in  §  69  that  0i,  .  .  .  ,  4>n 
show  the  same  number  of  variations  of  sign  for  x  =  r— p  as  for  x  =  r-\-p. 
The  same  is  true  for  /i,  .  .  .  ,  /„  since  the  products  of  </>i,  .  .  .  ,  ^„  by 
(x  —  r)m_1  have  for  a  given  x  the  same  signs  as  0i  .  .  .  ,  ^or  the  same 
signs  as  —fa,  ...  ,  —<fn-  But  the  latter  series  evidently  shows  the 
same  number  of  variations  of  sign  as  fa,  .  .  .  ,  </>M.  Hence  (4)  is  proved 
and  consequently  the  present  theorem. 

1  The  degree  of  f(x)  is  not  n,  nor  was  it  necessarily  n  in  §  69. 


73] 


BUDAN'S  THEOREM 


83 


EXERCISES 

1.  For/  =  x4-8x24-16,  prove  that  F1  =  x3-4x,  F2  =  x2-4,  Fx=xF2.  Hence  n  =  2. 
Verify  that  V-x—2,  Foo=0,  and  that  there  are  just  two  real  roots,  each  a  double 
root. 

Discuss  similarly  the  following  equations. 

2.  x4-5x3+9x2-7x+2  =  0.        3.  x4+2x3-3x2-4x+4  =  0.        4.  x4-x2-2x+2  =  0. 

73.  Budan's  Theorem.  Let  a  and  b  be  real  numbers,  a<b,  neither1 
a  root  of  f(x)=0,  an  equation  of  degree  n  with  real  coefficients.  Let  Va 
denote  the  number  of  variations  of  sign  of 

(12)  f(x),        f'(x),        f"{x),        ...,        /W(aO 

for  x  =  a,  after  vanishing  terms  have  been  deleted.  Then  Va—Vb  is  either 
the  number  of  real  roots  of  f(x)  =  0  between  a  and  b  or  exceeds  the  number 
of  those  roots  by  a  positive  even  integer.  A  root  of  multiplicity  m  is  here 
counted  as  m  roots. 

For  example,  if  /(x)  =x3-7x-7,  then  /'  =  3x2-7,  /"  =  6x,  /'"  =  6.  Their  values 
for  x  =  3,  4,  —2,  —1  are  tabulated  below. 


X 

/ 

/' 

/"    /'" 

Variations 

3 

-1 

20 

18     6 

1 

4 

29 

41 

24     6 

0 

-2 

-1 

5 

-12     6 

3 

-1 

-1 

-4 

-66 

1 

Hence  the  theorem  shows  that  there  is  a  single  real  root  between  3  and  4,  and  two 
or  no  real  roots  between  —2  and  —1.  The  theorem  does  not  tell  us  the  exact  number 
of  roots  between  the  latter  limits.  To  decide  this  ambiguity,  note  that  /( —3/2)  =  +1/8, 
so  that  there  is  a  single  real  root  between  —2  and  —1.5,  and  a  single  one  between 
—  1.5  and  —1. 

The  proof  is  quite  simple  if  no  term  of  the  series  (12)  vanishes  for 
x  =  a  or  for  x  =  6  and  if  no  two  consecutive  terms  vanish  for  the  same 
value  of  x  between  a  and  b.     Indeed,  if  no  one  of  the  terms  vanishes  for 


X\  2l  X  ^  X2, 

as  for  x  =  X2. 


then  VXl  =  VXv  since  any  term  has  the   same  sign   for  x  ■■ 


Xi 


Next,  let  r  be  a  root  of  fm(x)  =  0,  a<r<b.     By  hypothesis, 


1  In  case  a  or  b  is  a  root  of  /(x)  =0,  the  theorem  holds  if  we  count  the  number  of 
roots >o  and  =  b.  This  inclusive  theorem  has  been  proved,  by  means  of  Rolle's 
theorem,  by  A.  Hurwitz,  M athematische  Annalen,  Vol.  71,  1912,  p.  584,  who  extended 
Budan's  theorem  from  the  case  of  a  polynomial  to  a  function  /(x)  which  is  real  and 
regular  f or  a  ^  x  <  b . 


84  ISOLATION  OF  REAL  ROOTS  [Ch.  VI 

the  first  derivative  f(i+1)(x)  of  f{i)(x)  is  not  zero  for  x  =  r.  As  in  the  third 
step  (now  actually  the  case  4  =  0)  in  §  69,  f(i)(x)  and  fii+1)(x)  show  one 
more  variation  of  sign  for  x  —  r  —  p  than  for  x  =  r-\-p,  where  p  is  a  sufficiently 
small  positive  number.  If  i>0,  f(i)  is  preceded  by  a  term  /(i_1)  in  (12). 
By  hypothesis,  /w_1)(x)^0  for  x  —  r  and  hence  has  the  same  sign  for 
x  =  r  —  p  and  x  =  r-\-p  when  p  is  sufficiently  small.  For  these  values  of 
x,  fli)(x)  has  opposite  signs.  Hence /(i_1)  and /(i)  show  one  more  or  one  less 
variation  of  sign  for  x  =  r  —  p  than  for  x  =  r-\-p,  so  that/(i_1),  fH),  /(i+1)  show 
two  more  variations  or  the  same  number  of  variations  of  sign. 

Next,  let  no  term  of  the  series  (12)  vanish  for  x  =  a  or  for  x  =  6,  but 
let  several  sucessive  terms 

(13)  /»(*),        f«+1Kx),  ." .  •  ,  /»+'-»(*) 

all  vanish  for  a  value  r  of  x  between  a  and  6,  while  f{i+j){r)  is  not  zero, 
but  is  say  positive.1  Let  I\  be  the  interval  between  r  —  p  and  r,  and  I2 
the  interval  between  r  and  r-\-p.  Let  the  positive  number  p  be  so  small 
that  no  one  of  the  functions  (13)  or  f(i+j)(x)  is  zero  in  these  intervals,  so 
that  the  last  function  remains  positive.  Hence  f(i+}~l){x)  increases  with 
x  (since  its  derivative  is  positive)  and  is  therefore  negative  in  7i 
and  positive  in  I2.  Thus  fii+j~2)(x)  decreases  in  7i  and  increases  in  I2 
and  hence  is  positive  in  each  interval.  In  this  manner  we  may  verify  the 
signs  in  the  following  table: 

f(i)  /(*  +  !>  f(f+2)  ...  f(i+J-3)      f(i+j-2)     JH+j-l)        f{i+j) 

h    I  (->•  (-)'-1    (~>-2        •••  +  + 

h  I  +  +  +  ...+  +  +  + 

Hence  these  functions  show  j  variations  of  sign  in  7i  and  none  in  I2. 

If  i>0,  the  first  term  of  (13)  is  preceded  by  a  function  /(i_1)(x)  which 
is  not  zero  for  x  =  r,  and  hence  not  zero  in  7i  or  72  if  p  is  sufficiently  small. 

If  j  is  even,  the  signs  of  /w_1)  and  /(i)  are  H — h  or h  in  both  7i  and  72, 

showing  no  loss  in  the  number  of  variations  of  sign.  If  j  is  odd,  their 
signs  are 

7i        +  - 

or 
72        +  +  -  + 

so  that  there  is  a  loss  or  gain  of  a  single  variation  of  sign.     Hence 
f«-D         fa)         /(1+1),         •  f-i+J) 

1  If  negative,  all  signs  in  the  table  below  are  to  be  changed;  but  the  conclusion  holds 


§  73]  BU DAN'S  AND  DESCARTES'  THEOREMS  85 

show  a  loss  of  j  variations  of  sign  if  j  is  even,  and  a  loss  of  j±  1  if  j  is  odd, 
and  hence  always  a  loss  of  an  even  number  h  0  of  variations  of  sign. 

If  i  =  0,  fm  =f  has  r  as  a  j-f old  root  and  the  functions  in  the  table  show 
j  more  variations  of  sign  for  x  =  r  —  p  than  for  x  =  r-\-y. 

Thus,  when  no  one  of  the  functions  (12)  vanishes  for  x  =  a  or  for  x  =  b, 
the  theorem  follows  as  at  the  end  of  §  69  (with  unity  replaced  by  the 
multiplicity  of  a  root). 

Finally,  let  one  of  the  functions  (12),  other  than  f(x)  itself,  vanish  for 
x  =  a  or  for  x  =  6.  If  8  is  a  sufficiently  small  positive  number,  all  of  the 
N  roots  of  f(x)=0  between  a  and  b  lie  between  a-\-8  and  b—8,  and  for 
the  latter  values  no  one  of  the  functions  (12)  is  zero.  By  the  above 
proof, 

Va+5-VJ)-5  =  N+2t, 

Va-Va+S=2j,        Vb-S-Vb=2s, 

where  t,  j,  s  are  integers  2:  0.     Hence  Va—  Vb  =  N-{-2(t-\-j-\-s). 

Descartes'  rule  of  signs  (§  67)  is  a  corollary  to  Budan's  theorem.  Con- 
sider any  equation  with  real  coefficients 

f(x)=aoxn+aiXn~1+  .  .  .  -\-an_1x+an  =  0, 

having  a„^0.     For  x  —  0  the  functions  (12)  have  the  same  signs  as 

0-n,  On-l)  •  •  •  }  ai>  ao- 

Hence  Vq  is  equal  to  the  number  V  of  variations  of  sign  of  f(x) . 

For  x  =  +  oo ,  the  functions  all  have  the  same  sign,  which  is  that  of  ao. 
Thus  Fo —  F^oo  =  ^  is  either  the  number  of  positive  roots  or  exceeds  that 
number  by  a  positive  even  integer.  Finally,  Descartes'  rule  holds  if 
a„  =  0,  as  shown  by  removing  the  factors  x. 

EXERCISES 

Isolate  by  Budan's  theorem  the  real  roots  of 

1.  x3-x2-2x  +  l=0.  2.  x3+3x2-2a;-5=0. 

3.  Prove  that  if  f(a)  ?^0,  Va  equals  the  number  of  real  roots  >a  or  exceeds  that 
number  by  an  even  integer. 

4.  Prove  that  there  is  no  root  greater  than  a  number  making  each  of  the  functions 
(12)  positive,  if  the  leading  coefficient  of /(x)  is  positive.     (Newton.) 

5.  Hence  verify  that  xi  —  4x3  —  3x+23  =0  has  no  root  >4. 

6.  Show  that  x4— 4x3+x2+6x+2  =  0  has  no  root  >3. 


CHAPTER  VII 

Solution  of  Numerical  Equations 

74.  Horner's  Method.1  After  we  have  isolated  a  real  root  of  a  real 
equation  by  one  of  the  methods  in  Chapter  VI,  we  can  compute  the  root 
to  any  desired  number  of  decimal  places  either  by  Horner's  method, 
which  is  available  only  for  polynomial  equations,  or  by  Newton's  method 
(§75),  which  is  applicable  also  to  logarithmic,  trigonometric,  and  other 
equations. 

To  find  the  root  between  2  and  3  of 

(1)  x3- 2a:- 5  =  0, 

set  x  =  2-\-p.     Direct  substitution  gives  the  transformed  equation  for  p: 

(2)  p3+6p2+10p-l  =  0. 

The  method  just  used  is  laborious  especially  for  equations  of  high  degree. 
We  next  explain  a  simpler  method.     Since  p  =  x  —  2, 

x3-2x-5  =  (x-2)3+6(x-2)2  +  10(x-2)-l, 

identically  in  x.  Hence  —1  is  the  remainder  obtained  when  the  given 
polynomial  x3  —  2x  —  5  is  divided  by  x  —  2.     By  inspection,  the   quotient 

Q  is  equal  to 

O-2)2+6(:r-2)  +  10. 

Hence  10  is  the  remainder  obtained  when  Q  is  divided  by  x  —  2.  The 
new  quotient  is  equal  to  (x  —  2)-\-G,  and  another  division  gives  the 
remainder  6.  Hence  to  find  the  coefficients  6,  10,  —  1  of  the  terms  follow- 
ing p3  in  the  transformed  equation  (2),  we  have  only  to  divide  the  given 
polynomial  x3  —  2x—  5  by  z  — 2,  the  quotient  Q  by  x—2,  etc.,  and  take 
the  remainders  in  reverse  order.  However,  when  this  work  is  performed 
by  synthetic  division  (§  15)  as  tabulated  below,  no  reversal  of  order  is 

1  W.  G.  Horner,  London  Philosophical  Transactions,  1819.  Earlier  (1804)  by  P. 
Ruffini.     See  Bulletin  American  Math.  Society,  May,  1911. 

86 


74] 


HORNER'S  METHOD 


87 


necessary,  since  the  coefficients  then  appear  on  the  page  in  their  desired 
order. 

10-2-5     |2 
2  4  4 


1 

2 

2 

-1 

2 

8 

1 

4 
2 

10 

Thus  1,  6,  10,  —  1  are  the  coefficients  of  the  desired  equation  (2). 

To  obtain  an  approximation  to  the  decimal  p,  we  ignore  for  the  moment 
the  terms  involving  p3  and  p2;  then  by  lOp— 1  =  0,  p  =  0.1.  But  this 
value  is  too  large  since  the  terms  ignored  are  all  positive.  For  p  =  0.09, 
the  polynomial  in  (2)  is  found  to  be  negative,  while  for  p  =  0.1  it  was  just 
seen  to  be  positive.  Hence  p  =  0.09-\-h,  where  h  is  of  the  denomination 
thousandths.  The  coefficients  1,  6.27,  ...  of  the  transformed  equation 
for  h  appear  in  heavy  type  just  under  the  first  zigzag  line  in  the  following 
scheme: 

10.09 


0.05 
11.1 

=  0.004 


1    6 

0.09 

10 
0.5481 

-1 
0.949329 

1    6.09 
0.09 

10.5481 
0.5562 

-0.050671 

1    6.18 
0.09 

11.1043 

0.025096 

0.044517584 

1   6.27 

0.00 

4 

1    6.274 
0.004 

11.129396 
0.025112 

-0.006153416 

1    6.278 
0.004 

11 . 154508 

1    6.28 

2 

Hence  z  =  2.094+^  where  t  is  a  root  of 

*3+6.282Z2+ll.  154508*- 0.006153416  =  0. 
By  the  last  two  terms,  t  is  between  0.0005  and  0.0006.     Then  the  value 


88  SOLUTION  OF  NUMERICAL  EQUATIONS  [Ch.  VII 

of  C=t3 +6.282t2  is  found  to  lie  between  0.00000157  and  0.00000227. 
Hence  we  may  ignore  C  provided  the  constant  term  be  reduced  by  an 
amount  between  these  limits.  Whichever  of  the  two  limits  we  use,  we 
obtain  the  same  dividend  below  correct  to  6  decimal  places. 

11.154508  |     0.006151     |  0.000551=1 

5577 


574 
558 

16 
11 


Since  the  quotient  is  0.0005 +,  only  two  decimal  places  of  the  divisor  are 
used,  except  to  see  by  inspection  how  much  is  to  be  carried  when  making 
the  first  multiplication.  Hence  we  mark  a  cross  above  the  figure  5  in 
the  hundredths  place  of  the  divisor  and  use  only  11.15.  Before  making 
the  multiplication  by  the  second  significant  figure  5  of  the  quotient  t, 
we  mark  a  cross  over  the  figure  1  in  the  tenths  place  of  the  divisor  and 
hence  use  only  11.1.  Thus  x  =  2.0945514+ ,  with  doubt  only  as  to  whether 
the  last  figure  should  be  4  or  5. 

If  we  require  a  greater  number  of  decimal  places,  it  is  not  necessary 
to  go  back  and  construct  a  new  transformed  equation  from  the  equation 
in  t.  We  have  only  to  revise  our  preceding  dividend  on  the  basis  of  our 
present  better  value  of  t.  We  now  know  that  t  is  between  0.000551  and 
0.000552.  To  compute  the  new  value  of  the  correction  C,  in  which  we 
may  evidently  ignore  ts,  we  use  logarithms. 

log      5.51   =    .74115  log      5.52  =    .74194 

/.  log      5 . 5 12  =  1 .  48230  .'.  log      5 .  522  =  1 .  48388 

log      6.282=    .79810  log      6.282=    .79810 


log  190.72   =2.28040  log  191.42   =2.28198 

Hence  C  is  between  0.000001907  and  0.000001915.  Whichever  of  the  two 
limits  we  use,  we  obtain  the  same  new  dividend  below  correct  to  8  decimal 
places. 


§74] 


HORNER'S  METHOD 


X    X  XX  X 

11.154508 


0.00615150 
557725 


57425 
55773 

1652 
1115 


537 
446 

91 

89 


0.00055148 


LIBRARY  SEISM0L0GIC 
OBSERVATORY 


Hence,  finally,  x  =  2.094551482,  with  doubt  only  as  to  the  last  figure. 

EXERCISES 

(The  number  of  transformations  made  by  synthetic  division  should  be  about  half 
the  number  of  significant  figures  desired  for  a  root.) 

By  one  of  the  methods  in  Chapter  VI,  isolate  each  real  root  of  the  following  equa- 
tions, and  compute  each  real  root  to  5  decimal  places. 


1.  x3 +2x4-20  =  0. 
3.  x3+x2-2x-l=0. 
5.  x4 - ll,727x+40,385  = 


=0. 


2.  x3+3x2-2x-5  =  0. 

4.  x4+4x3-17.5x2-l8x-f-58.5=0. 

6.  x3  =  10. 


Find  to  7  decimal  places  all  the  real  roots  of 

7.  x3+4x2-7  =  0.  8.  x3-7x-7  =  0. 

Find  to  8  decimal  places 

9.  The  root  between  2  and  3  of  x3  —  x— 9  =  0  (make  only  3  transformations). 

10.  The  real  cube  root  of  7.976. 

11.  The  abscissa  of  the  real  point  of  intersection  of  the  conies  y  =  x2,  xy+x+Sy  — 
6  =  0. 

12.  Find  to  3  decimal  places  the  abscissas  of  the  points  of  intersection  of  x2+y2  =  9, 
y  =  x2  —  x. 

13.  A  sphere  two  feet  in  diameter  is  formed  of  a  kind  of  wood  a  cubic  foot  of  which 
weighs  two-thirds  as  much  as  a  cubic  foot  of  water  (i.e.,  the  specific  gravity  of  the  wood 
is  2/3).  Find  to  four  significant  figures  the  depth  h  to  which  the  floating  sphere 
will  sink  in  water. 

Hints:  The  volume  of  a  sphere  of  radius  r  is  ^vr3.     Hence  our  sphere  whose  radius 


90  SOLUTION  OF  NUMERICAL  EQUATIONS  [Ch.  VII 

is  1  foot  weighs  as  much  as  ■§*'"§■  cubic  feet  of  water.  The  volume  of  the  submerged 
portion  of  the  sphere  is  Trh2(r—^h)  cubic  feet.  Since  this  is  also  the  volume  of  the  dis- 
placed water,  its  value  for  r  =  \  must  equal  -g-n--f      Hence  h3  —  3/i2+§  =  0. 

14.  If  the  specific  gravity  of  cork  is  1/4,  find  to  four  significant  figures  how  far  a 
cork  sphere  two  feet  in  diameter  will  sink  in  water. 

15.  Compute  cos  20°  to  four  decimal  places  by  use  of 

cos  SA  =4  cosM.—  3  cos  A,        cos  60°  =  -g-. 

16.  Three  intersecting  edges  of  a  rectangular  parallelopiped  are  of  lengths  6,  8, 
and  10  feet.  If  the  volume  is  increased  by  300  cubic  feet  by  equal  elongations  of  the 
edges,  find  the  elongation  to  three  decimal  places. 

17.  Given  that  the  volume  of  a  right  circular  cylinder  is  aw  and  the  total  area  of 
its  surface  is  2j3tt,  prove  that  the  radius  r  of  its  base  is  a  root  of  r3  — /3r+a  =  0.  If  a  =  56, 
(3  =  28,  find  to  four  decimal  places  the  two  positive  roots  r.  The  corresponding  altitude 
is  a/r2. 

18.  What  rate  of  interest  is  implied  in  an  offer  to  sell  a  house  for  $2700  cash,  or 
in  annual  installments  each  of  $1000  payable  1,  2,  and  3  years  from  date? 

Hint:  The  amount  of  $2700  with  interest  for  3  years  should  be  equal  to  the  sum 
of  the  first  payment  with  interest  for  2  years,  the  amount  of  the  second  payment  with 
interest  for  1  year,  and  the  third  payment.  Hence  if  r  is  the  rate  of  interest  and  we 
write  x  for  1  +r,  we  have 

2700  x3  =  1000  x2+1000  a; +1000. 

19.  Find  the  rate  of  interest  implied  in  an  offer  to  sell  a  house  for  $3500  cash,  or  in 
annual  installments  each  of  $1000  payable  1,  2,  3,  and  4  years  from  date. 

20.  Find  the  rate  of  interest  implied  in  an  offer  to  sell  a  house  for  $3500  cash,  or 
$4000  payable  in  annual  installments  each  of  $1000,  the  first  payable  now. 

75.  Newton's  Method.  Prior  to  1676,  Newton  1  had  already  found 
the  root  between  2  and  3  of  equation  (1).  He  replaced  x  by  2+p  and 
obtained  (2).  Since  p  is  a  decimal,  he  neglected  the  terms  in  pz  and  p2, 
and  hence  obtained  p  =  0.1,  approximately.  Replacing  p  by  0.1+g  in 
(2),  he  obtained 

g3+6.3g2+11.23g+0.061  =  0. 

Dividing  —0.061  by  11.23,  he  obtained  —0.0054  as  the  approximate 
value  of  q.     Neglecting  q3  and  replacing  q  by  —  0.0054+r,  he  obtained 

6.3r2  + 11. 16196r+0.000541708  =  0. 

Dropping  6.3r2,  he  found  r  and  hence 

x  =  2+0. 1  -  0.0054  -  0.00004853  =  2.09455147, 

1  Isaac  Newton,  Opuscula,  I,  1794,  p.  10,  p.  37. 


75] 


NEWTON'S  METHOD 


91 


of  which  all  figures  but  the  last  are  correct  (§  74).  But  the  method  will 
not  often  lead  so  quickly  to  so  accurate  a  value  of  the  root. 

Newton  used  the  close  approximation  0.1  to  p,  in  spite  of  the  fact 
that  this  value  exceeds  the  root  p  and  hence  led  to  a  negative  correction 
at  the  next  step.  This  is  in  contrast  with  Horner's  method  in  which  each 
correction  is  positive,  so  that  each  approximation  must  be  chosen  less 
than  the  root,  as  0.09  for  p. 

Newton's  method  may  be  presented  in  the  following  general  form, 
which  is  applicable  to  any  equation  f(x)  =  0,  whether  f(x)  is  a  polynomial 
or  not.  Given  an  approximate  value  a  of  a  real  root,  we  can  usually 
find  a  closer  approximation  a+h  to  the  root  by  neglecting  the  powers 
h2,  h3,  .  .  .  of  the  small  number  h  in  Taylor's  formula  (§  56) 


f(a+h)  =f(a)+f(a)h+f"(a)^+ 


and  hence  by  taking 


f(a)+f'(a)h  =  0, 


h  = 


-f(fl) 

/'(a)  ' 


We  then  repeat  the  process  with  ai  =  a-\-h  in  place  of  the  former  a. 
Thus  in  Newton's  example,  f(x)  =xz  —  2x  —  5,  we  have,  for  a  =  2, 

-/(2)_1 
/'(2)       10' 


h- 


ai  =  a-\-h  =  2.l, 


hi  = 


-/(2.1)     -0.061 


/'(2-1) 


11.23 


=  -0.0054 


76.  Graphical  Discussion  of  Newton's  Method.  Using  rectangular 
coordinates,  consider  the  graph  of  y=f(x)  and  the  point  P  on  it  with  the 
abscissa  OQ  =  a  (Fig.  22).     Let  the  tangent  at  P  meet  the  a>axis  at  T 


Fig.  22 


Fig.  23 


92 


SOLUTION  OF  NUMERICAL  EQUATIONS 


[Ch.  VII 


and  let  the  graph  meet  the  z-axis  at  S.     Take  h  =  QT,  the  subtangent. 
Then 

~f(a) 


QP=f(a),        /'(a)=tanX7T  =  - 


h    ' 


h  = 


-f(a) 
f'(a)  " 


In  the  graph  in  Fig.  22,  OT  =  a-\-h  is  a  better  approximation  to  the 
root  OS  than  OQ  =  a.  The  next  step  (indicated  by  dotted  lines)  gives  a 
still  better  approximation  OT\. 

If,  however,  we  had  begun  with  the  abscissa  a  of  a  point  Pi  in  Fig.  22 
near  a  bend  point,  the  subtangent  would  be  very  large  and  the  method 
would  probably  fail  to  give  a  better  approximation.  Failure  is  certain 
if  we  use  a  point  P2  such  that  a  single  bend  point  lies  between  it  and  S. 

We  are  concerned  with  the  approximation  to  a  root  previously  isolated 
as  the  only  real  root  between  two  given  numbers  a  and  8.  These  should 
be  chosen  so  nearly  equal  that  f'(x)  =0  has  no  real  root  between  a  and  8, 
and  hence/(a;)  =  y  has  no  bend  point  between  a  and  8.  Further,  if  f"(x)  =  0 
has  a  root  between  our  limits,  our  graph  will  have  an  inflexion  point  with 
an  abscissa  between  a  and  8,  and  the  method  will  likely  fail  (Fig.  23). 

Let,  therefore,  neither  f'(x)  nor  f"(x)  vanish  between  a  and  /3.  Since 
/"  preserves  its  sign  in  the  interval  from  a  to  /3,  while  /  changes  in  sign, 
/"  and  /  will  have  the  same  sign  for  one  end  point.  According  as  the 
abscissa  of  this  point  is  a  or  (3,  we  take  a  =  a  or  a  =  8  for  the  first  step  of 
Newton's  process.  In  fact,  the  tangent  at  one  of  the  end  points  meets 
the  x-axis  at  a  point  T  with  an  abscissa  within  the  interval  from  a.  to  8. 
If  f'(x)  is  positive  in  the  interval,  so  that  the  tangent  makes  an  acute 
angle  with  the  z-axis,  we  have  Fig.  24  or  Fig.  25;  if  f  is  negative,  Fig. 
26  or  Fig.  22. 


«    T 


Fig.  24 


Fig.  25 


Fig.  26 


§  76]  GRAPHICAL  DISCUSSION  OF  NEWTON'S  METHOD  93 

In  Newton's  example,  the  graph  between  the  points  with  the  abscissas  a  =  2  and 
/3  =  3  is  of  the  type  in  Fig.  24,  but  more  nearly  like  a  vertical  straight  line.  In  view 
of  this  feature  of  the  graph,  we  may  safely  take  a=a,  as  did  Newton,  although  our 
general  procedure  would  be  to  take  a  =/3.  The  next  step,  however,  accords  with  our 
present  process;  we  have  a  =  2,  0  =  2.1  in  Fig.  24  and  hence  we  now  take  a  =  /3,  getting 

0.061 

=  0.0054 

11  23 

as  the  sub  tangent,  and  hence  2.1  —0.0054  as  the  approximate  root. 

If  we  have  secured  (as  in  Fig.  24  or  Fig.  26)  a  better  upper  limit  to  the 
root  than  /3,  we  may  take  the  abscissa  c  of  the  intersection  of  the  chord 
A B  with  the  a>axis  as  a  better  lower  limit  than  a.     By  similar  triangles, 

-f(a)  :c-a=m  :  (3-c, 
whence 

a/(/3)-/3/(a) 


(3) 


/G8)-/(«) 


This  method  of  finding  the  value  of  c  intermediate  to  a  and  /3  is  called  the 
method  of  interpolation  {regula  falsi) . 

In  Newton's  example,  a  =  2,  /3  =  2.1, 

/(«)  =  -l,        /(,3)  =0.061,         c  =  2.0942. 

The  advantage  of  having  c  at  each  step  is  that  we  know  a  close  limit 
of  the  error  made  in  the  approximation  to  the  root. 

We  may  combine  the  various  possible  cases  discussed  into  one: 

If  f(x)  =  0  has  a  single  real  root  between  a  and  /3,  and  f'{x)  =0,  f"{x)  =  0 
have  no  real  root  between  a  and  /3,  and  if  we  designate  by  /3  that  one  of  the 
numbers  a  and  fi  for  which  /(/3)  and  f"{0)  have  the  same  sign,  then  the  root 
lies  in  the  narrower  interval  from  c  to  0—  f(0)/f'(0),  where  c  is  given  by  (3). 

It  is  possible  to  prove  l  this  theorem  algebraically  and  to  show  that  by 
repeated  applications  of  it  we  can  obtain  two  limits  a',  {$'  between  which 
the  root  lies,  such  that  a'—fi'  is  numerically  less  than  any  assigned  posi- 
tive number.  Hence  the  root  can  be  found  in  this  manner  to  any  desired 
accuracy. 

Examine.        f(x)  =x3-2x2-2,        a  =  2\,         P=2%.     Then 

f(a)  =  -U,  /(£)=!■ 

^Weber's  Algebra,  2d  ed.,  I,  pp.  380-382;  Kleines Lehrbuch  der  Algebra,  1912,  p.  163. 


94  SOLUTION  OF  NUMERICAL  EQUATIONS  [Ch.  VII 

Neither  of  the  roots  0,  4/3  of  fix)  =  0  lies  between  a  and  ft  so  that  /(a;)  =0  has  a  single 
real  root  between  these  limits  (§  65) .  Nor  is  the  root  f  of  fix)  =  0  within  these  limits. 
The  conditions  of  the  theorem  are  therefore  satisfied.  For  a<z</3,  the  graph  is  of 
the  type  in  Fig,  24,    We  find  that  approximately 

c=Mf  =  2.3487,        ft  =  0-^  =  2.3714, 

ft-^T  =  2.3597. 

/'(ft) 

For  a;  =  2.3593,  f(x)=  -0.00003.  We  therefore  have  the  root  to  four  decimal  places 
For  a  =  2.3593, 

/'(a)  =7.2620,        a-^  =2.3593041, 
/(a) 

which  is  the  value  of  the  root  correct  to  7  decimal  places.  We  at  once  verify  that  the 
result  is  greater  than  the  root  in  view  of  our  work  and  Fig.  24,  while  if  we  change  the 
final  digit  from  1  to  0,  fix)  is  negative. 

EXERCISES 

1.  For  f(x)=xi+x3-Sx2-x-A,  show  by  Descartes'  rule  of  signs  that/'(x)=0 
and  f"(x)  =0  each  have  a  single  positive  root  and  that  neither  has  a  root  between  1 
and  2.     Which  of  the  values  1  and  2  should  be  taken  as  ft? 

2.  When  seeking  a  root  between  2  and  3  of  x3-x-9  =  0,  which  value  should  be 
taken  as  ft? 

77.  Systematic  Computation  of  Roots  by  Newton's  Method.     By  way 

of  illustration  we  shall  compute  to  7  decimal  places  a  positive  root  of 

f(x)  =  x4 + x3  -  3x2  -  x  -  4  =  0. 
Since  /(l)  =  —  6,  /(2)  =  6,  there  is  a  real  root  between  1  and  2.     Since 
/'Or)  =4^+3^-6^-1,        /'(1)=0,        /'(2)=31, 

the  graph  ofy=f(x)  is  approximately  horizontal  near  (1,  —6)  and  approxi- 
mately vertical  near  (2,  6).  Hence  the  root  is  much  nearer  to  2  than  to  1. 
Thus  in  applying  Newton's  method  we  employ  a  =  2  as  the  first  approxi- 
mation to  the  root.     The  correction  h  is  then 

-/(2)_-6_ 

The  work  of  performing  the  substitutions  x  =  2+d,  d=  —  0.2-f-e,  ..., 
to  find  the  transformed  equations  satisfied  by  d,  e,  .  .  .  ,  is  done  by  syn- 


§77] 


COMPUTATION  OF  ROOTS  BY  NEWTON'S  METHOD 


95 


thetic  division,  exactly  as  in  Horner's  method,  except  that  some  of  the 
multipliers  are  now  negative: 

1  1 


8.2 

0.04 


8.16 
-0.04 


8.12 
0.04 


8.08 
0.04 


8.04 


-3 

6 


-1 
6 


0.3264 


—0.860544 


21.5136 
-0.3248 


20.387456 
-0.847552 


21 . 1888 
-0.3232 


19.539904 


20.8656 


-4 
10 


1 

3 

3 

5 

6 

2 

10 

26 

1 

5 
2 

13 
14 

31 

1 

7 

27 

2 

-0.2 


-0.2 

-1.76 

-5.048 

-5.1904 

8.8 
-0.2 

25.24 
-1.72 

25 . 952 
-4.704 

0.8096 

8.6 
-0.2 

23.52 

-1.68 

21.248 

8.4 
-0.2 

21.84 

-0.8096 
21.248 
=  -0.04 


■0.81549824 


-0.00589824 


0.005898 
19.54 


=  0.000302 


The  root  is  2-0.2-0.04+0.000302=1.760302,  in  which  the  last 
figure  is  in  slight  doubt.  Indeed,  it  can  be  proved  that  if  the  final  fraction 
g,  when  expressed  as  a  decimal,  has  k  zeros  between  the  decimal  point  and  the 
first  significant  figure,  the  division  may  be  safely  carried  to  2k  decimal  places. 
In  our  example  k  —  S,  so  that  we  retained  6  decimal  places  in  g. 

To  proceed  independently  of  this  rule,  we  note  that  g  is  obviously 
between  0.00030  and  0.00031.     Then  the  value  of  g4 + 8  Mg3 +20. 8Q5Qg2 


96  SOLUTION  OF  NUMERICAL  EQUATIONS  [Ch.  VII 

is  found  to  lie  between  0.000001878  and  0.000002006.  Whichever  of  these 
limits  we  use  as  a  correction  by  which  to  reduce  the  constant  term,  we 
obtain  the  same  dividend  below  correct  to  6  decimal  places. 


XX      XX 


19.539904  |  0.005896  |  0.0003017 
005862 


34 
20 

14 

14 

Hence  the  root  is  1.7603017  to  7  decimal  places. 

EXERCISES 

1.  Find  to  8  decimal  places  the  root  between  2  and  3  of  z3-x-9  =  0. 

2.  Find  to  7  decimal  places  the  root  between  2  and  3  of  x3  —  2x2  — 2  =  0. 

3.  Find  the  real  cube  root  of  7.976  to  5  decimal  places. 

4.  Explain  by  Taylor's  expansion  of /(2-',-d)  why  the  values  of 

/(2),    /m    ire),    ^f"'w>    dr4r"(2) 

are  in  reverse  order  the  coefficients  of  the  transformed  equation 

d4+9d3+27d2+31d+6  =  0, 
obtained  in  the  Example  in  the  text,  and  printed  in  heavy  type. 

5.  The  method  commonly  used  to  find  the  positive  square  root  of  n  by  a  computing 
machine  consists  in  dividing  n  by  an  assumed  approximate  value  a  of  the  square  root 
and  taking  half  the  sum  of  a  and  the  quotient  as  a  better  approximation.  Show  that 
the  latter  agrees  with  the  value  of  a+h  given  by  applying  Newton's  method  to 
f(x)  =x2—n. 

78.  Newton's  Method  for  Functions  not  Polynomials. 

Example  1 .  Find  the  angle  x  at  the  center  of  a  circle  subtended  by  a  chord  which 
cuts  off  a  segment  whose  area  is  one-eighth  of  that  of  the  circle. 

Solution.  If  x  is  measured  in  radians  and  if  r  is  the  radius,  the  area  of  the  segment 
is  equal  to  the  left  member  of 

^r 2  (x — sin  x)  =  -girr2, 
whence 

x  —  sin  £  =  -j7r. 


§  78]    NEWTON'S    METHOD    FOR    FUNCTIONS    NOT    POLYNOMIALS    97 

By  means  of  a  graph  of  y  =  smx  and  the  straight  line  represented  by  y=x  — £n-,  we 
see  that  the  abscissa  of  their  point  of  intersection  is  approximately  1.78  radians  or  102°. 
Thus  a  =  102°  is  a  first  approximation  to  the  root  of 

/(x)  =x— sin  x— ^tt  =  0. 

By  Newton's  method  a  better  approximation  is  a-\-h,  where  * 

—f(a)      — a+sina+T7r 


h  = 


f'(a)  1—  cos  a 


sin  102°  =  0.9781  cos  102°  =  -0.2079 

£(3.1416)=  0.7854  1- cos  102°=  1.2079 

1.7635  -0.0167 

h  = =  -0.0138 

102  °  =  1 .7802  radians  1 .2079 


-0.0167 


Oi  =  o+ft  =  1.7664 


L       -/(ai)      -1.7664+0.9809+0.7854 

/li  = = =  —  U  .UUU1 . 

/'(a:)  1.1944 

Hence  x  =  ai+hi  =  1.7663  radians,  or  101°  12' 

Example  2.2     Solve  x— log  x  =  7,  the  logarithm  being  to  base  10. 

Solution.  Evidently  x  exceeds  7  by  a  positive  decimal  which  is  the  value  of  log  x. 
Hence  in  a  table  of  common  logarithms,  we  seek  a  number  x  between  7  and  8  whose 
logarithm  coincides  approximately  with  the  decimal  part  of  x.  We  read  off  the  values 
in  the  second  column. 


7.897 
7.898 


log  x 


0.89746 
0 . 89752 


x— log  X 


6.99954 
7.00048 


By  the  final  column  the  ratio  of  interpolation  is  46/94.     Hence  a;  =  7.8975  to  four 
decimal  places. 

1  The  derivative  of  sin  x  is  cos  x.     We  need  the  limit  of 

sin  (x+2k)  —sin  x  _2  cos  ^(2x+2k)  sin  -g-(2/c)  _cos  (x+k)  sin  k 
2k  2k  k 

as  2k  approaches  zero.     Since  the  ratio  of  sin  k  to  k  approaches  1,  the  limit  is  cos  x. 

2  This  Ex.  2,  which  should  be  contrasted  with  Ex.  3,  is  solved  by  interpolation 
since  that  method  is  simpler  than  Newton's  method  in  this  special  case. 


98  SOLUTION  OF  NUMERICAL  EQUATIONS  (Ch.  Vll 

Example  3.     Solve  2x  —  log  x  =  7f  the  logarithm  being  to  base  10. 
Solution.    Evidently  x  is  a  little  less  than  4.    A  table  of  common  logarithms  shows 
at  once  that  a  fair  approximation  to  x  is  a  =  3.8.     Write 

/(x)=2x-logx-7,        log  x=ilf  loge  x,        M  =  0.4343. 
By  calculus,  the  derivative  of  loge  x  is  1/x.    Hence 

f(x)=2-— ,        /'(a)  =2-0.1143  =  1.8857, 

x 

/(a)  =0.6-log  3.8  =  0.6-0.57978  =  0.02022, 

_/j =4^=0.0107,        ai  =  a+ft  =  3.7893, 

/(ai)  =0.000041,        /(3.7892)  =  -0.000148. 

—  X0.0001=  0.000078,        x=3.789278. 
189 

All  figures  of  x  are  correct  as  shown  by  Vega's  table  of  logarithms  to  10  places. 

EXERCISES 

Fine?  the  angle  x  at  the  center  of  a  circle  subtended  by  a  chord  which  cuts  off  a  seg- 
ment whose  ratio  to  the  circle  is 

1.  i-  2.  f. 

When  the  logarithms  are  to  base  10, 

3.  Solve2x-logx  =  9.  4.  Solve  3x-log  x  =  9. 

5.  Find  the  angle  just  >  15°  for  which  ^  sin  x  +sin  2x  =  0.64. 

6.  Find  the  angle  just  >  72°  for  which  x— |  sin  x  =  \ir. 

7.  Find  all  solutions  of  Ex.  5  by  replacing  sin  2x  by  2  sin  a;  cos  x,  squaring,  and 
solving  the  quartic  equation  for  cos  x. 

8.  Solve  similarly  sin  x+sin  2x  =  1.2. 

9.  Find  x  to  6  decimal  places  in  sin  x  =  x  —  2. 
10.  Find  x  to  5  decimal  places  in  x  =  3  loge  x. 

79.  Imaginary  Roots.  To  find  the  imaginary  roots  x-\-yi  of  an  equa- 
tion /(V)=0  with  real  coefficients,  expand  f(x-\-yi)  by  Taylor's  theorem; 
we  get 

/(x)+fd)!/t-/"W14-/'"WI|L3+  •  •  •  =0. 


79]  IMAGINARY  ROOTS  99 

Since  x  and  y  are  to  be  real,  and  y^O, 


(4) 


/(^)-/,/(^)^+r,,wr^4- . . .  =o, 
f(X)-r\z)j3^f<*(x)£- ...  =o. 


In  the  Example  and  Exercises  below,  f(z)  is  of  degree  4  or  less.  Then 
the  second  equation  (4)  is  linear  in  y2.  Substituting  the  resulting  value 
of  y2  in  the  first  equation  (4),  we  obtain  an  equation  E(x)  =  0,  whose  real 
roots  may  be  found  by  one  of  the  preceding  methods.  If  the  degree  of 
f(z)  exceeds  4,  we  may  find  E(x)  —  0  by  eliminating  y2  between  the  two 
equations  (4)  by  one  of  the  methods  to  be  explained  in  Chapter  X. 


Example.     For  f(z)  =  z4  — 2+1,  equations  (4)  are 

z4-x+l-6xy+2/4  =  0,        4x3-l-4:n/2=0. 


Thus 


2/2=3.2-—  _4a.6_l_3.2_L.        =Q. 

Q.X  10 


The  cubic  equation  in  x2  has  the  single  real  root 

x2  =  0.528727,         x  =  ±0.72714. 
Then  y2  =  0.184912  or  0.87254,  and 

2  =  x+_yi-0.72714±0.43001i,         -0.72714±0.93409i. 

EXERCISES 

Find  the  imaginary  roots  of 

1.  23-2z-5=0.  2.  28z3+922-l=0. 

3.  z4-3z2-62  =  2.  4.  z4-4s3+llz2-14z+10<-0. 

5.  z4-4z3+9z2-16z+20  =  0.     Hint: 

#(x)=x(x-2)(16x4-64a-3  +  136:c2-144x+65)=0, 

and  the  last  factor  becomes  (w2 -\-l) (w2 -\-9)  for  2x  =  w+2. 

Note.  If  we  know  a  real  root  r  of  a  cubic  equation  /(z)  =0,  we  may  remove  the 
factor  z—r  and  solve  the  resulting  quadratic  equation.  When,  as  usual,  r  involves 
several  decimal  places,  this  method  is  laborious  and  unsatisfactory.  But  we  may  utilize 
a  device,  explained  in  the  author's  Elementary  Theory  of  Equations,  pp.  119-121,  §§  6,  7. 
As  there  explained,  a  simila.  device  may  be  used  when  we  know  two  real  roots  of  a 
quartic  equation. 


100  SOLUTION  OF  NUMERICAL  EQUATIONS  [Ch.  VII 

MISCELLANEOUS  EXERCISES 

(Give  answers  to  6  decimal  places,  unless  the  contrary  is  stated.) 

1.  What  arc  of  a  circle  is  double  its  chord? 

2.  What  arc  of  a  circle  is  double  the  distance  from  the  center  of  the  circle  to  the 
chord  of  the  arc? 

3.  If  A  and  B  are  the  points  of  contact  of  two  tangents  to  a  circle  of  radius  unity 
from  a  point  P  without  it,  and  if  arc  AB  is  equal  to  PA,  find  the  length  of  the  arc. 

4.  Find  the  angle  at  the  center  of  a  circle  of  a  sector  which  is  bisected  by  its  chord. 

5.  Find  the  radius  of  the  smallest  hollow  iron  sphere,  with  air  exhausted,  which  will 
float  in  water  if  its  shell  is  1  inch  thick  and  the  specific  gravity  of  iron  is  7.5. 

6.  From  one  end  of  a  diameter  of  a  circle  draw  a  chord  which  bisects  the  semicircle. 

7.  The  equation  xtan:r=c  occurs  in  the  theory  of  vibrating  strings.  Its  approxi- 
mate solutions  may  be  found  from  the  graphs  of  y  =  cot  x,  y  =  x/c.     Find  x  when  c  =  1. 

8.  The  equation  tan  x=x  occurs  in  the  study  of  the  vibrations  of  air  in  a  spherical 
cavity.  From  an  approximate  solution  Xi  =  1.5ir,  we  obtain  successively  better  approxi- 
mations a;2=tan-1a;i  =  1.4334  7T,  x3  =  tan~1  x2,  ....  Find  the  first  three  solutions  to 
4  decimal  places. 

9.  Find  to  3  decimal  places  the  first  five  solutions  of 

2x 
tan  x  = 


2-x2' 


which  occurs  in  the  theory  of  vibrations  in  a  conical  pipe. 

10.  4tx3  —  (3x  — 1)2  =  0  arises  in  the  study  of  the  isothermals  of  a  gas.     Find   its 
roots  when  (i)  T  =  0.002  and  (ii)  t=0.99. 

11.  Solve  xx  =  100.  12.  Solve  x  =  10  log  x.  13.  Solve  x +log  x  =  x  log  a;. 

14.  Solve  Kepler's  equation  M  =  x-e  sin  x  when  M  =  332°  28'  54.8",  e  =  14°  3'  20". 

15.  In  what  time  would  a  sum  of  money  at  6%  interest  compounded  annually 
amount  to  as  much  as  the  same  sum  at  simple  interest  at  8%? 

16.  In  a  semicircle  of  diameter  x  is  inscribed  a  quadrilateral  with  sides  a,  b,  c,  x; 
then  x*-(a2+b2+c2)x-2abc  =  0  (I.  Newton).     Given  a  =  2,  6  =  3,  c  =  4,  find  x. 

17.  What  rate  of  interest  is  implied  in  an  offer  to  sell  a  house  for  $9000  cash,  or 
$1000  down  and  $3000  at  the  end  of  each  year  for  three  years? 


CHAPTER  VIII 


Determinants;  Systems  of  Linear  Equations 

80.  Solution  of  Two  Linear  Equations  by  Determinants  of  Order  2. 

Assume  that  there  is  a  pair  of  numbers  x  and  y  for  which 


(1) 


aix+biy  =  ki, 

a2X+b2y  =  k2. 


Multiply  the  members  of  the  first  equation  by  b2  and  those  of  the  second 
equation  by  —61,  and  add  the  resulting  equations.     We  get 

(ai&2  —  a2b\)X  —  &1&2  —  k2b\. 

Employing  the  respective  multipliers  —  a2  and  a\,  we  get 

(a\b2  —  o2bi)y  =  a\k2  —  a2k\. 
The  common  multiplier  of  x  and  y  is 
(2)  a\b2  —  a2b\, 

and  is  denoted  by  the  symbol 


(2') 


a2  b2 


which  is  called  a  determinant  of  the  second  order,  and  also  called  the  deter- 
minant of  the  coefficients  of  x  and  y  in  equations  (1).  The  results  above 
may  now  be  written  in  the  form 


(3) 


a\  bx 
a2  62 


k\  b\ 
k2  b2 


a\  b\ 
a2  b2 


y- 


a\  k\ 
a2  k2 


We  shall  call  k\  and  k2  the  known  terms  of  our  equations  (1).  Hence, 
if  D  is  the  determinant  of  the  coefficients  of  the  unknowns,  the  product  of  D  by 
any  one  of  the  unknowns  is  equal  to  the  determinant  obtained  from  D  by 
substituting  the  known  terms  in  place  of  the  coefficients  of  that  unknown. 

101 


102  DETERMINANTS;  SYSTEMS  OF  LINEAR  EQUATIONS    [Ch.  VIII 

If  D 5^0,  relations  (3)  uniquely  determine  values  of  x  and  y: 


x  = 


kib2  —  k2bi 


U 


a\k,2  —  a,2ki 


D  D 

and  these  values  satisfy  equations  (1);  for  example, 

aix+biy= y> =  *i- 

Hence  our  equations  (1)  have  been  solved  by  determinants  when  D^O. 
We  shall  treat  in  §  96  the  more  troublesome  case  in  which  D  =  0. 


Example.     For  2x  —  3y=  —  4, 


2 

-3 

-4  -3 

6 

-2 

X  — 

2   -2 

1 

4y  = 

2   -4 
6       2 

6x— 2y  =  2,  we  have 

14a;  =  14, 

=  28,  y=2. 


x  =  l, 


EXERCISES 

Solve  by  determinants  the  following  systems  of  equations: 

1.  8x-2/  =  34,         2.  3z+4t/  =  10,         3.  az+fo/  =  a2, 
x+8y  =  53.  4x+  y=  9.  bx  —  ay  =  ab. 

81.  Solution  of  Three  Linear  Equations  by  Determinants  of  Order  3. 

Consider  a  system  of  three  linear  equations 

aix+biy-\-ciz  =  ki, 

(4)  a,2X-\-b2y-\-C2z  =  k2, 

azx-\-bzy-irC2,z  =  kz. 

Multiply  the  members  of  the  first,  second  and  third  equations  by 

(5)  &2C3  —  bzC2,         &3C1  —  61C3,         b\C2  —  bza, 

respectively,  and  add  the  resulting  equations.  We  obtain  an  equation 
in  which  the  coefficients  of  y  and  z  are  found  to  be  zero,  while  the  coeffi- 
cient of  x  is 

(6)  ai&2C3  — ai&3C2  +  «2&3Cl— a2&lC3  +  a3&lC2  — a3&2Cl. 


82] 


SIGNS  OF  TERMS  OF  A  DETERMINANT 


103 


Such  an  expression  is  called  a  determinant  of  the  third  order  and  denoted 
by  the  symbol 

I  cii  b\  c\ 

(6')  |   0,2   62    C2 

I   «3   03   C3 

The  nine  numbers  a-i,  .  .  .  ,  C3  are  called  the  elements  of  the  determi- 
nant. In  the  symbol  these  elements  lie  in  three  (horizontal)  rows,  and 
also  in  three  (vertical)  columns.  Thus  0,2,  62,  C2  are  the  elements  of  the 
second  row,  while  the  three  c's  are  the  elements  of  the  third  column. 

The  equation  (free  of  y  and  z),  obtained  above,  may  now  be  written 
as 


a,\  b\  c\ 

fci  b\  C\ 

&2   02   C2 

x  — 

Ji2    62    C2 

03  63  C3 

&3    63    C3 

since  the  right  member  was  the  sum  of  the  products  of  the  expressions 
(5)  by  k\,  /c2,  &3,  and  hence  may  be  derived  from  (6)  by  replacing  the 
a's  by  the  /c's.  Thus  the  theorem  of  §  80  holds  here  as  regards  the 
unknown  x.  We  shall  later  prove,  without  the  laborious  computations 
just  employed,  that  the  theorem  holds  for  all  three  unknowns. 

82.  The  Signs  of  the  Terms  of  a  Determinant  of  Order  3.     In  the 

six  terms  of  our  determinant  (6),  the  letters  a,  b,  c  were  always  written 
in  this  sequence,  while  the  subscripts  are  the  six  possible  arrangements 
of  the  numbers  1,  2,  3.  The  first  term  ai&2C3  shall  be  called  the  diagonal 
term,  since  it  is  the  product  of  the  elements  in  the  main  diagonal  running 
from  the  upper  left-hand  corner  to  the  lower  right-hand  corner  of  the 
symbol  (6')  for  the  determinant.  The  subscripts  in  the  term  —a\bzC2 
are  derived  from  those  of  the  diagonal  term  by  interchanging  2  and  3, 
and  the  minus  sign  is  to  be  associated  with  the  fact  that  an  odd  number 
(here  one)  of  interchanges  of  subscripts  were  used.  To  obtain  the  arrange- 
ment 2,  3,  1  of  the  subscripts  in  the  term  -\-a2b3C1  from  the  natural  order 
1,  2,  3  (in  the  diagonal  term),  we  may  first  interchange  1  and  2,  obtaining 
the  arrangement  2,  1,  3,  and  then  interchange  1  and  3;  an  even  number 
(two)  of  interchanges  of  subscripts  were  used  and  the  sign  of  the  term 
is  plus. 

While  the  arrangement  1,  3,  2  was  obtained  from  1,  2,  3  by  one  inter- 
change (2,  3),  we  may  obtain  it  by  applying  in  succession  the  three  inter- 


104  DETERMINANTS;   SYSTEMS  OF  LINEAR  EQUATIONS  [Ch.  VIII 

changes  (1,  2),  (1,  3),  (1,  2),  and  in  many  new  ways.  To  show  that  the 
number  of  interchanges  which  will  produce  the  final  arrangement  1,  3,  2 
is  odd  in  every  case,  note  that  each  of  the  three  possible  interchanges, 
viz.,  (1,  2),  (1,  3),  and  (2,  3),  changes  the  sign  of  the  product 

P=(xi  —  x2)  (xi  -  xs.)  (x2  —  x3) , 

where  the  x's  are  arbitrary  variables.  Thus  a  succession  of  k  interchanges 
yields  P  or  —  P  according  as  k  is  even  or  odd.  Starting  with  the  arrange- 
ment 1,  2,  3  and  applying  k  successive  interchanges,  suppose  that  we 
obtain  the  final  arrangement  1,  3,  2.  But  if  in  P  we  replace  the  subscripts 
1,  2,  3  by  1,  3,  2,  respectively,  i.e.,  if  we  interchange  2  and  3,  we  obtain 
—P.  Hence  k  is  odd.  We  have  therefore  proved  the  following  rule 
of  signs: 

Although  the  arrangement  r,  s,  t  of  the  subscripts  in  any  term  ±ajbsct  of 
the  determinant  may  be  obtained  from  the  arrangement  1,  2,  3  by  various 
successions  of  interchanges,  the  number  of  these  interchanges  is  either  always 
an  even  number  and  then  the  sign  of  the  term  is  plus  or  always  an  odd  num- 
ber and  then  the  sign  of  the  term  is  minus. 

EXERCISES 

Apply  the  rule  of  signs  to  all  terms  of 

1.  Determinant  (6).  2.  Determinant  oib2— ct2?>i. 

83.  Number  of  Interchanges  always  Even  or  always  Odd.     We  now 

extend  the  result  in  §  82  to  the  case  of  n  variables  xi,  .  .  .  ,  xtt.  The 
product  of  all  of  their  differences  Xi  —  Xj(i<j)  is 

P=(xi  —  x2)(xi  —  x3)  .  .  .  (xi—  xa) 

■{X2  —  a&)  •  •  •  (x2  —  xa) 


'  \%n  —  1      •£»/  • 

Interchange  any  two  subscripts  i  and  j.  The  factors  which  involve  neither 
i  nor  j  are  unaltered.  The  factor  (xt—Xj)  involving  both  is  changed  in 
sign.     The  remaining  factors  may  be  paired  to  form  the  products 

zk(xi—xt)(xj—xt)         (k  =  l,  .  .  .  ,  n;     k^i,  k^j). 

Such  a  product  is  unaltered.     Hence  P  is  changed  in  sign. 

Suppose  that  an  arrangement  i\,  i2,   •  .  .  ,  in  can  be  obtained  from 


84] 


DEFINITION  OF  A  DETERMINANT  OF  ORDER  n 


105 


1,  2,  .  .  .  ,  n  by  using  m  successive  interchanges  and  also  by  t  successive 
interchanges.  Make  these  interchanges  on  the  subscripts  in  P;  the 
resulting  functions  are  equal  to  (  —  l)mP  and  (—  1)'P,  respectively.  But 
the  resulting  functions  are  identical  since  either  can  be  obtained  at  one 
step  from  P  by  replacing  the  subscript  1  by  i\,  2  by  12,  .  .  .  ,  n  by  ia.     Hence 

(-l)mp=(-l)'P, 

so  that  m  and  t  are  both  even  or  both  odd. 

Thus  if  the  same  arrangement  is  derived  from  1,2,  ...  ,nby  m  successive 
interchanges  as  by  t  successive  interchanges,  then  m  and  t  are  both  even  or 
both  odd. 

84.  Definition  of  a  Determinant  of  Order  n.  We  define  a  determinant 
of  order  4  to  be 


(7) 


ai  6i 

c\  d\ 

a2  62 

C2    d2 

a3  63 

cs  dz 

a±  64 

C4,  d± 

=  2ii  ±a<ibrCsdt, 

(24) 


where  q,  r,  s,  t  is  any  one  of  the  24  arrangements  of  1,  2,  3,  4,  and  the 
sign  of  the  corresponding  term  is  +  or  —  according  as  an  even  or  odd 
number  of  interchanges  are  needed  to  derive  this  arrangement  q,  r,  s,  t 
from  1,  2,  3,  4.  Although  different  numbers  of  interchanges  will  produce 
the  same  arrangement  q,  r,  s,  t  from  1,  2,  3,  4,  these  numbers  are  all  even 
or  all  odd,  as  just  proved,  so  that  the  sign  is  fully  determined. 

We  have  seen  that  the  analogous  definitions  of  determinants  of  orders 
2  and  3  lead  to  our  earlier  expressions  (2)  and  (6). 

We  will  have  no  difficulty  in  extending  the  definition  to  a  determinant 
of  general  order  n  as  soon  as  we  decide  upon  a  proper  notation  for  the  n2 
elements.  The  subscripts  1,  2,  .  .  .  ,  n  may  be  used  as  before  to  specify 
the  rows.  But  the  alphabet  does  not  contain  n  letters  with  which  to 
specify  the  columns.  The  use  of  e',  e",  .  .  .  ,  e(re)  for  this  purpose  would 
conflict  with  the  notation  for  derivatives  and  besides  be  very  awkward 
when  exponents  are  used.  It  is  customary  in  mathematical  journals 
and  scientific  books  (a  custom  not  always  followed  in  introductory  text 
books,  to  the  distinct  disadvantage  of  the  reader)  to  denote  the  n  letters 
used  to  distinguish  the  n  columns  by  ei,  eo,  •  •  •  ,  en  (or  some  other  letter 
with  the  same  subscripts)  and  to  prefix  (but  see  §  85)  such  a  subscript  by 


106 


DETERMINANTS;  SYSTEMS  OF  LINEAR  EQUATIONS    [Ch.  VIII 


the  new  subscript  indicating  the  row.     The  symbol  for  the  determinant 

is  therefore 

en  ei2  .  .  .  ei„ 


(8) 


D  = 


621    622    ...    62» 


2  •  1  terms 


By  definition  this  shall  mean  the  sum  of  the  n(n—  1) 

(9)  (-iyehiei22  .  .  .  etnn 

in  which  i\,  12,  .  .  .  ,  4  is  an  arrangement  of  1,  2,  .  .  .  ,  n,  derived  from 
1,  2,  .  .  .  ,  n  by  i  interchanges.  Any  term  (9)  of  the  determinant  (8)  is, 
apart  from  sign,  the  product  of  n  factors,  one  and  only  one  from  each  col- 
umn, and  one  and  only  one  from  each  row. 

For  example,  if  we  take  n=4  and  write  ah  bh  ch  dj  for  eji,  e&,  ^3,  %4, 
the  symbol  (8)  becomes  (7)  and  the  general  term  (9)  becomes  the  general 
term  (—1)*  ah  bi2  ciz  du  of  the  second  member  of  (7) . 

EXERCISES 

1.  Find  the  six  terms  involving  a^  in  the  determinant  (7). 

2.  What  are  the  signs  of  G^sC^ie^  aibiddkey  in  a  determinant  of  order  five? 

3.  Show  that  the  arrangement  4,  1,  3,  2  may  be  obtained  from  1,  2,  3,  4  by  use  of 
the  two  successive  interchanges  (1,  4),  (1,  2),  and  also  by  use  of  the  four  successive 
interchanges  (1,  4),  (1,  3),  (1,  2),  (2,  3). 

4.  Write  out  the  six  terms  of  (8)  for  n  =  3,  rearrange  the  factors  of  each  term  so  that 
the  new  first  subscripts  shall  be  in  the  order  1,  2,  3,  and  verify  that  the  resulting  six 
terms  are  those  of  the  determinant  D'  in  §  85  for  ra  =  3. 

85.  Interchange  of  Rows  and  Columns.  Any  determinant  is  not 
altered  in  value  if  in  its  symbol  we  replace  the  elements  of  the  first,  second, 
.  .  .  ,  nth  rows  by  the  elements  which  formerly  appeared  in  the  same  order 
in  the  first,  second,  .  .  .  ,  nth  columns,  or  briefly  if  we  interchange  the  cor- 
responding rows  and  columns.     For  example, 

a  c 

b  d 
We  are  to  prove  that  the  determinant  D  given  by  (8)  is  equal  to 

en  e2\   ...  en\ 


a  b 
c  d 


=  ad  —  bc  = 


D'  = 


e\2    ^22 


en2 


eln     e2n 


§  85]  INTERCHANGE  OF  ROWS  AND  COLUMNS  107 

If  we  give  to  D'  a  more  familiar  aspect  by  writing  eilc  =  au  for  each  element 
so  that,  as  in  (8),  the  row  subscript  precedes  instead  of  follows  the  column 
subscript,  the  definition  of  the  determinant  in  terms  of  the  a's  gives  D' 
in  terms  of  the  e's  as  the  sum  of  all  expressions 

(  —  l)leu1e2£2-  •  •  enbn, 

in  which  k\,  &2  •  •  .  ,  kn  is  an  arrangement  of  1,  2,  .  .  .  ,  n,  derived  from 
the  latter  sequence  by  i  interchanges. 

As  for  the  terms  of  D,  without  altering  (9),  we  may  rearrange  its  factors 
so  that  the  first  subscripts  shall  appear  in  the  order  1,  2,  .  .  .  ,  n,  and 
obtain 

(—1)^1^2*2  •   •   •  enkn- 

This  can  be  done  by  performing  in  reverse  order  the  i  successive  inter- 
changes of  the  letters  e  corresponding  to  the  i  successive  interchanges 
which  were  used  to  derive  the  arrangement  i\,  i%,  .  .  .  ,  in  of  the  first 
subscripts  from  the  arrangement  1,  2,  .  .  .  ,  n.  Thus  the  new  second 
subscripts  ki,  .  .  .  ,  kn  are  derived  from  the  old  second  subscripts  1,  .  .  .  , 
n  by  i  interchanges.  The  resulting  signed  product  is  therefore  a  term 
ofD'.     Hence  D  =  D'. 

86.  Interchange  of  Two  Columns.  A  determinant  is  merely  changed 
in  sign  by  the  interchange  of  any  two  of  its  columns.     For  example, 


D  = 


a  b 
c  d 


=  ad  —  bc,         A  = 


b  a 
d  c 


=  bc—ad=  —D. 


Let  A  be  the  determinant  derived  from  (8)  by  the  interchange  of  the 
rth  and  sth  columns.  The  terms  of  A  are  therefore  obtained  from  the 
terms  (9)  of  D  by  interchanging  r  and  s  in  the  series  of  second  subscripts. 
Interchange  the  rth  and  sth  letters  e  to  restore  the  second  subscripts 
to  their  natural  order.  Since  the  first  subscripts  have  undergone  an 
interchange,  the  negative  of  any  term  of  A  is  a  term  of  D,  and  A=  —  D. 

87.  Interchange  of  Two  Rows.  A  determinant  D  is  merely  changed 
in  sign  by  the  interchange  of  any  two  rows. 

Let  A  be  the  determinant  obtained  from  D  by  interchanging  the  rth 
and  .sth  rows.  By  interchanging  the  rows  and  columns  in  D  and  in  A, 
we  get  two  determinants  D'  and  A',  either  of  which  may  be  derived  from 
the  other  by  the  interchange  of  the  rth  and  sth  columns.  Hence,  by 
§§  85,  86, 

A  =  A'=-D'=-D. 


108 


DETERMINANTS;    SYSTEMS  OF  LINEAR  EQUATIONS    [Ch.  VIII 


88.  Two  Rows  or  Two  Columns  Alike.  A  determinant  is  zero  if  any 
two  of  its  rows  or  any  two  of  its  columns  are  alike. 

For,  by  the  interchange  of  the  two  like  rows  or  two  like  columns,  the 
determinant  is  evidently  unaltered,  and  yet  must  change  in  sign  by  §§  86, 
87.     Hence  D=-D,D  =  0. 

EXERCISES 


1.  Prove  that  the  equation  of  the  straight  line  determined  by  the  two  distinct 
points  (xi,  yi)  and  (x2,  y2)  is 

x     y      1 

xi    yi     1    =0. 

£2    2/2     1 

2.  Show  that 
ai    61    Ci 

a%    b2    c-2 

a3     h     cz 

By  use  of  the  Factor  Theorem  (§  14)  and  the  diagonal  term,  prove  that 


a% 

C2 

&2 

Ol 

Cl 

h 

= 

a3 

c3 

63 

«3 

Ol 

a% 

h 

61 

b2 

Cz 

Cl 

c2 

1    1    1 

a      b      c 

a2     b2    c2 

1  .  . 

x2  . 
X22  . 


=  (b— a){c— a)(c— 6). 


1 

xn 

■  %n 


Xi 


n-l 


X2 


n-l 


Xn 


=  n  (xi—xj). 

i,j=l 


This  is  known  as  the  determinant  of  Vandermonde,  who  discussed  it  in  1770.     The 
symbol  on  the  right  means  the  product  of  all  factors  of  the  type  indicated. 
5.  Prove  that  a  skew-symmetric  determinant  of  odd  order  is  zero: 


=  0. 


0 

a 

b 

c 

d 

0 

a 

b 

—a 

0 

e 

f 

9 

—a 

0 

c 

=0, 

-b 

— e 

0 

h 

3 

-b 

— c 

0 

—c 

-f 

-  h 

0 

k 

-d 

-9 

—i 

-* 

0 

§90] 


EXPANSION  BY  ROW  OR  COLUMN 


109 


a2 

c2 

,     B2  = 

ai 

Ci 

,      £3  = 

Cli 

Ci 

a3 

c3 

«3 

c3 

CLi 

c-z 

89.  Minors.  The  determinant  of  order  n—\  obtained  by  erasing 
(or  covering  up)  the  row  and  column  crossing  at  a  given  element  of  a 
determinant  of  order  n  is  called  the  minor  of  that  element. 

For  example,  in  the  determinant  (6')  of  order  3,  the  minors  of  &i,  b2,  63  are  respect- 
ively 

B,= 

Again,  (6')  is  the  minor  of  di  in  the  determinant  of  order  4  given  by  (7). 

90.  Expansion  According  to  the  Elements  of  a  Row  or  Column.     In 

a\     b\     c\ 
(6')  D=      a2    b2    c2 

0,3       &3       C3 

denote  the  minor  of  any  element  by  the  corresponding  capital  letter, 
so  that  &i  has  the  minor  B\,  b%  has  the  minor  B3,  etc.,  as  in  §  89.  We 
shall  prove  that 


D=     aiAi-biBi+ciCi, 

D=  —0,2A2  +  b2B2  —  C2C2, 

D=     a3A3  —  b3B3 +C3C3, 


D=     aiAi-a2A2+a3A3, 
D=  -biBl+b2B2-b3B3, 
D=     c1Ci-c2C2+c3C3. 


The  three  relations  at  the  left  (or  right)  are  expressed  in  words  by  saying 
that  a  determinant  D  of  the  third  order  may  be  expanded  according  to  the 
elements  of  the  first,  second  or  third  row  (or  column) .  To  obtain  the  expan- 
sion, we  multiply  each  element  of  the  row  (or  column)  by  the  minor  of 
the  element,  prefix  the  proper  sign  to  the  product,  and  add  the  signed 
products.     The  signs  are  alternately  +  and  — ,  as  in  the  diagram 

+  -  + 
-  +  - 
+      -      + 

For  example,  by  expansion  according  to  the  second  column, 


-4X9= -36. 


1     4     5 

2     3 

2     0     3 

=  _4 

3     9 

3     0     9 

110         DETERMINANTS;    SYSTEMS  OF  LINEAR  EQUATIONS   [Ch.  VIII 


Similarly  the  value  of  the  determinant  (7)  of  order  4  may  be  found  by  expansion 
according  to  the  elements  of  the  fourth  column : 


■4 


We  shall  now  prove  that  any  determinant  D  of  order  n  may  be  expanded 
according  to  the  elements  of  any  row  or  any  column. 

Let  Eij  denote  the  minor  of  etj  in  D,  given  by  (8),  so  that  Eti  is 
obtained  by  erasing  the  ith.  row  and  jth  column  of  D. 

(i)  We  first  prove  that 


a2      &2      C2 

ai    bi    d 

ai   bi    C\ 

«i    &i 

C! 

a3    bz    c3 

+d> 

a-i    h    c3 

-d, 

a-i    b2    c2 

+di 

a2    bi 

c2 

a\   hi    Ct 

0,4   bi   Ci 

^    bi    Ci 

a3    63 

c3 

(10) 


D  =  enEn-e2iE2i+e31E31-  .  .  .  +(-l)w-1e„1#r, 


so  that  D  may  be  expanded  according  to  the  elements  of  its  first  column. 
By  (9)  the  terms  of  D  having  the  factor  en  are  of  the  form 

(-l)4ene*22.  .  .  ev, 

where  1,  12,  •  ■  ■  ,  in  is  an  arrangement  of  1,  2,  .  .  .  ,  n,  obtained  from  the 
latter  by  i  interchanges,  so  that  12,  •  •  •  ,  in  is  an  arrangement  of  2,  ...  , 
n,  derived  from  the  latter  by  i  interchanges.  After  removing  from  each 
term  the  common  factor  en  and  adding  the  quotients,  we  obtain  a  sum 
which,  by  definition,  is  the  value  of  the  determinant  En  of  order  n—1. 
Hence  the  terms  of  D  having  the  factor  en  may  all  be  combined  into 
en  En,  which  is  the  first  part  of  (10). 

We  next  prove  that  the  terms  of  D  having  the  factor  621  may  be  com- 
bined into  —  621-^21,  which  is  the  second  part  of  (10).  For,  if  A  be  the 
determinant  obtained  from  D  by  interchanging  its  first  and  second  rows, 
the  result  just  proved  shows  that  the  terms  of  A  having  the  factor  621 
may  be  combined  into  the  product  of  e2i  by  the  minor 


612  ei3 

^32    633 


esn 


of  621  in  A.      Now  this  minor  is  identical  with  the  minor  #21  of  e2i  in  D. 
But  A=  —  D  (§  87).     Hence  the  terms  of  D  having  the  factor  e2i  may  be 


91] 


REMOVAL  OF  FACTORS 


111 


combined  into  — 621-^21-     Similarly,  the  terms  of  D  having  the  factor  631 
may  be  combined  into  e^iE^i,  etc.,  as  in  (10). 

(ii)  We  next  prove  that  D  may  be  expanded  according  to  the  elements 
of  its  kth.  column  (fc>l): 


(11) 


d=s  (-iy+hjbEjt. 


Consider  the  determinant  8  derived  from  D  by  moving  the  kth  column 
over  the  earlier  columns  until  it  becomes  the  new  first  column.  Since 
this  may  be  done  by  k  —  1  interchanges  of  adjacent  columns,  5  =  (—  l)i_1D. 
The  minors  of  the  elements  elk,  .  .  .  ,  enb  in  the  first  column  of  8  are  evidently 
the  minors  Elk,  .  .  .  ,  Enk  of  elk,  .  .  .  ,  enk  in  D.     Hence,  by  (10), 

8  =  elkElk-e2kE2k+ +(-l)n-1enkEnk=  t  (-iy+lejkEjk. 

3=1 

Thus  D=(-l)k-18  has  the  desired  value  (11). 

(m)  Finally,  D  may  be  expanded  according  to  the  elements  of  its 
kth  row: 

D=t  (-iy+%jEkJ. 

In  fact,  by  Case  (ii) ,  the  latter  is  the  expansion  of  the  equal  determinant 
D'  in  §  85  according  to  the  elements  of  its  &th  column. 

91.  Removal  of  Factors.  A  common  factor  of  all  of  the  elements  of  the 
same  row  or  same  column  of  a  determinant  may  be  divided  out  of  the  elements 
and  placed  as  a  factor  before  the  new  determinant. 

In  other  words,  if  all  of  the  elements  of  a  row  or  column  are  divided 
by  n,  the  value  of  the  determinant  is  divided  by  n.     For  example, 


na\   nb\ 

—  n 

ai  61 

02       62 

a<z  62 

a\  nb\  c\ 

a\  b\  c\ 

a<z  n&2  C2 

=n 

02    62    C2 

a%  nbz  c3 

03  63  c3 

Proof  is  made  by  expanding  the  determinants  according  to  the  elements 
of  the  row  or  column  in  question  and  noting  that  the  minors  are  the  same 
for  the  two  determinants.     Thus  the  second  equation  is  equivalent  to 

-{nbl)Bx  +  (nb2)B2-(nbz)B-s  =  n{-blBl+b2B2-b-iBz), 

where  Bx  denotes  the  minor  of  6*  in  the  final  determinant. 


112  DETERMINANTS;  SYSTEMS  OF  LINEAR  EQUATIONS     [Ch.  VIII 

EXERCISES 

2r     I     2>r 

=  0. 


1. 

3o    36     3c 

2. 

2r     I     3r 

5a    5b     5c 

=  0. 

2s     m   3s 

d      e      f 

2t     n    3t 

by  the  shortest  method  and  evaluate 

3. 

2    7    3 

4. 

5    7    0 

5    9     8 

6    8    0 

0    3     0 

3    9    4 

a 

b 

c     d 

a2 

b* 

c2    d2 

a3 

b3 

c3    d3 

a4 

b4 

c4    d4 

=  abcd(a  — 6)  (a— c)(a— d)(6— c)(&— d)(c— a"). 


92.  Sum  of  Determinants.  A  determinant  having  ai+gi,  02+^2,  ...  as 
f/ie  elements  of  a  column  is  equal  to  the  sum  of  the  determinant  having  a\, 
a2,  ...  as  the  elements  of  the  corresponding  column  and  the  determinant 
having  q\,  q2,  ...  as  the  elements  of  that  column,  while  the  elements  of  the 
remaining  columns  of  each  determinant  are  the  same  as  in  the  given  determi- 
nant. 

For  example, 

ai+qi       61     c\ 

«2  +  ?2         b2      C2 

a3~\-q3       03     C3 

To  prove  the  theorem  we  have  only  to  expand  the  three  determinants 
according  to  the  elements  of  the  column  in  question  (the  first  column  in 
the  example)  and  note  that  the  minors  are  the  same  for  all  three  determi- 
nants. Hence  ai+gi  is  multiplied  by  the  same  minor  that  a\  and  q\ 
are  multiplied  by  separately,  and  similarly  for  £12+02,  etc. 

The  similar  theorem  concerning  the  splitting  of  the  elements  of  any 
row  into  two  parts  is  proved  by  expanding  the  three  determinants  accord- 
ing to  the  elements  of  the  row  in  question.     For  example, 


a\  61  c\ 

a2  62  C2 

+ 

CI3    &3    C3 

QX 

61 

C\ 

Q2 

02 

C2 

C/3 

63 

C3 

a-\-r 


b+s 
d 


a  b 

r  s 

= 

+ 

c  d 

c  d 

93] 


ADDITION  OF  COLUMNS  OR  ROWS 


113 


93.  Addition  of  Columns   or  Rows.     A   determinant  is  not  changed 

in  value  if  we  add  to  the  elements  of  any  column  the  products  of  the  correspond- 
ing elements  of  another  column  by  the  same  arbitrary  number. 

Let  ai,  a,2,  .  .  .  be  the  elements  to  which  we  add  the  products  of  the 
elements  &i,  &2,  ...  by  n.  We  apply  §92  with  qi=nb\,  q2  =  nb2,  .... 
Thus  the  modified  determinant  is  equal  to  the  sum  of  the  initial  determi- 
nant and  a  determinant  having  61,  62,  ...  in  one  column  and  nb\,  nb2, 
...  in  another  column.  But  (§  91)  the  latter  determinant  is  equal  to 
the  product  of  n  by  a  determinant  with  two  columns  alike  and  hence 
is  zero  (§  88).     For  example, 


ai-\-nbi     61     c\ 

a\  b\  c\ 

hi  b\  c\ 

<22  +  n&2       &2       C2 

= 

02  62  C2 

-\-n 

62  62  C2 

a3+nbs     63     c3 

as  63  c3 

&3    63    C3 

and  the  last  determinant  is  zero. 

Similarly,  a  determinant  is  not  changed  in  value  if  we  add  to  the  elements 
of  any  row  the  products  of  the  corresponding  elements  of  another  row  by  the 
same  arbitrary  number. 

For  example, 


-\-nc 

b+nd 

a    b 

+n 

c  d 

a  b 

c 

d 

c    d 

c  d 

c  d 

Example.     Evaluate  the  first  determinant  below. 


1 

-2     1 

1 

2     3 

= 

6 

4     3 

1 

0     1 

1 

8     3 

= 

6 

10    3 

0 

0     1 

-2 

8  1 

2 

8     3 

= 

3 

10  1 

3 

10    3 

-44. 


Solution.  First  we  add  to  the  elements  of  the  second  column  the  products  of  the 
elements  of  the  last  column  by  2.  In  the  resulting  second  determinant,  we  add  to  the 
elements  of  the  first  column  the  products  of  the  elements  of  the  third  column  by  —  1 . 
Finally,  we  expand  the  resulting  third  determinant  according  to  the  elements  of  its 
first  row, 


1.  Prove  that 


EXERCISES 


b  -\-c  c  +a  a  -j-b 
bi+ci  ci+ai  ai+bi 
62+C2     C2+32     a2+b2 


a 

b 

c 

Oi 

h 

Ci 

di 

62 

Cl 

114 


DETERMINANTS;   SYSTEMS  OF  LINEAR  EQUATIONS    [Ch.  VIII 


By  reducing  to  a  determinant  of  order  3,  etc.,  prove  that 


2. 


3. 


1 

1 

1      1 

a 

b 

c      d 

a2 

b2 

c2    d2 

a3 

¥ 

c3    d3 

2 

-1 

3  -2 

1 

7 

1   -1 

3 

5  - 

-5      3 

4 

-3 

2  -1 

=  (a-b)(a-c)(a-d)(b-c)(b-d)(c-d). 


-42. 


1  1 

1  2 

1  3 

1  4 


1 

1 

3 

4 

6 

10 

10 

20 

=  1. 


94.  System  of  n  Linear  Equations  in  n  Unknowns  with  ZM0.     In 

a\iXi-\-ai2X2+  .  .  .  +alnxn  =  ki, 


(12) 


dn\X\-\- ttn2X2 ~T    •   •   •    "r^mj^Ti  —  ™»> 

let  Z)  denote  the  determinant  of  the  coefficients  of  the  n  unknowns : 

an  ai2  .  .  .  aln 


D  = 


Then 


Dxi 


anl     an2 


anxi  ai2  . 

.  d\n 

anxi+ai2a;2+  . 

•     i^ln^ny 

ai2  • 

.  dj 

(lnlXl      0,n2    • 

•   •   dnn 

anlxi-\-an2x2+  . 

•    ~\^nn^ni 

an2  • 

•  a„ 

where  the  second  determinant  was  derived  from  the  first  by  adding  to 
the  elements  of  the  first  column  the  products  of  the  corresponding  elements 
of  the  second  column  by  X2,  etc.,  and  finally  the  products  of  the  elements 
of  the  last  column  by  xn.  The  elements  of  the  new  first  column  are  equal 
to  fci,  .  .  .  ,  fc„  by  (12).     In  this  manner,  we  find  that 

(13)  Dxi=Ki,        Dx2  =  K2,         ...,        Dxn  =  Kn, 

in  which  Kt  is  derived  from  D  by  substituting  k\,  .  .  .  ,  kn  for  the  elements 
au,  ...,  ani  of  the  ^th  column  of  D,  whence 

k\  a\2  .  .  .  aln 
Ki  = ,  .  .  .  ,        Kn  = 


kn  a„ 


■  a„ 


an  ...  ain_!  ki 
anl   .  .  .  ann_!  kn 


94] 


EQUATIONS  WITH  DETERMINANT  NDT  ZERO 


115 


If  ZMO,  the  unique  values  of  xi,  .  .  .  ,  xn  determined  by  division 
from  (13)  actually  satisfy  equations  (12).  For  instance,  the  first  equation 
is  satisfied  since 

ki     an     a\2  •  .  •  a-, 


k\D  —  a\\K\  —  CI12K2  — 


—  a\nKn  — 


fci 

k2 


an 
a2i 


«12 
«22 


aln 

a2n 


Kn    af{ 


<(, 


and 


as  shown  by  expansion  according  to  the  elements  of  the  first  row; 
the  determinant  is  zero,  having  two  rows  alike. 

Theorem.  If  D  denotes  the  determinant  of  the  coefficients  of  the  n 
unknowns  in  a  system  of  n  linear  equations,  the  product  of  D  by  any  one 
of  the  unknowns  is  equal  to  the  determinant  obtained  from  D  by  substituting 
the  known  terms  in  place  of  the  coefficients  of  that  unknown.  If  D^O,  we 
obtain  the  unique  values  of  the  unknowns  by  division  by  D. 

We  have  therefore  given  a  complete  proof  of  the  results  stated  and 
illustrated  in  §  80,  §  81.  Another  proof  is  suggested  in  Ex.  7  below. 
The  theorem  was  discovered  by  induction  in  1750  by  G.  Cramer. 


EXERCISES 

Solve  by  determinants  the  following  systems  of  equations  (reducing  each  deter- 
minant to  one  having  zero  as  the  value  of  every  element  but  one  in  a  row  or  column, 
as  in  the  example  in  §  93) . 


1.     x  +  y+  3  =  11, 

2x  —  Qy—  2  =  0, 
3z+42/+2z  =  0. 

3.     x-2y+  z  =  12, 

x +2y +32  =  48, 

6x+4?/+3z  =  84. 

5.  x+  y-\-  2+  w=  1, 
x+2y+  32+  4u>  =  ll, 
x+Zy+  62+10w  =  26, 
x+4?/+102+20w;  =  47. 


2.  x+  y+  2  =  0, 
x+2y+3z=-l, 
x+3y+Gz  =  0. 

4.  Bx-2y  =  7, 
3y-2z  =  6, 
3z-2x=-l. 

6.  2x-  y+3z-2w  =  i, 

x+7y+  2—  w  —  2, 

3x+5j/-53+3to=0, 

4x  —  3y+2z—  w  =  5. 


7.  Prove  the  first  relation  (13)  by  multiplying  the  members  of  the  first  equation 
(12)  by  An,  those  of  the  second  equation  by  — A2i,  .  .  .  ,  those  of  the  nth  equation  by 
(  — l)n-1A„i,  and  adding,  where  Ay  denotes  the  minor  of  ay  in  D.  Hint:  The  resulting 
coefficient  of  x2  is  the  expansion,  according  to  the  elements  of  its  first  column,  of  a  deter- 
minant derived  from  D  by  replacing  an  by  ai2)  .  .  .  ,  onl  by  an2. 


116  DETERMINANTS;   SYSTEMS  OF  LINEAR  EQUATIONS    [Ch.VIII 

95.  Rank  of  a  Determinant.  If  we  erase  from  a  determinant  D  of 
order  n  all  but  r  rows  and  all  but  r  columns,  we  obtain  a  determinant 
of  order  r  called  an  r-rowed  minor  of  D.  In  particular,  any  element  is 
regarded  as  a  one-rowed  minor,  and  D  itself  is  regarded  as  an  n-rowed 
minor. 

If  a  determinant  D  of  order  n  is  not  zero,  it  is  said  to  be  of  rank  n. 
If,  for  0<r<n,  some  r-rowed  minor  of  D  is  not  zero,  while  every  (r+1)- 
rowed  minor  is  zero,  D  is  said  to  be  of  rank  r.  It  is  said  to  be  of  rank 
zero  if  every  element  is  zero. 

For  example,  a  determinant  D  of  order  3  is  of  rank  3  if  D^O;  of  rank  2  if  D  —  0, 
but  some  two-rowed  minor  is  not  zero;  of  rank  1  if  every  two -rowed  minor  is  zero, 
but  some  element  is  not  zero.    Again,  every  three-rowed  minor  of 


a 

b 

c 

d 

e 

f 

9 

h 

a 

b 

c 

d 

e 

f 

g 

h 

is  zero  since  two  pairs  of  its  rows  are  alike.  Hence  it  is  of  rank  2  if  some  two-rowed 
minor  is  not  zero.  But  it  is  of  rank  1  if  a,  b,  c,  d  are  not  all  zero  and  are  proportional 
to  e,  /,  g,  h,   since  all  two-rowed  minors  are  then  zero. 

96.  System  of  n  Linear  Equations  in  n  Unknowns  with  D  =  0.    We 

shall  now  discuss  the  equations  (12)  for  the  troublesome  case  (previously 
ignored)  in  which  the  determinant  D  of  the  coefficients  of  the  unknowns 
is  zero.  In  view  of  (13),  the  given  equations  are  evidently  inconsistent 
if  any  one  of  the  determinants  Ki,  .  .  .  ,  Kn  is  not  zero.  But  if  D  and 
these  K's  are  all  zero,  our  former  results  (13)  give  us  no  information 
concerning  the  unknowns  xi}  and  we  resort  to  the  following 

Theorem.  Let  the  determinant  D  of  the  coefficients  of  the  unknowns 
in  equations  (12)  be  of  rank  r,  r<n.  If  the  determinants  K  obtained  from 
the  (r-\-l)-rowed  minor t>  of  D  by  replacing  the  elements  of  any  column  by 
the  corresponding  known  terms  ki  are  not  all  zero,  the  equations  are  incon- 
sistent. But  if  these  determinants  K  are  all  zero,  the  r  equations  involving 
the  elements  of  a  non-vanishing  r-rowed  minor  of  D  determine  uniquely  r 
of  the  unknowns  as  linear  functions  of  the  remaining  n  —  r  unknowns,  which 
are  independent  variables,  and  the  expressions  for  these  r  unknowns  satisfy 
also  the  remaining  n  —  r  equations. 


96] 


EQUATIONS  WITH  DETERMINANT  ZERO 


117 


Consider  for  example  the  three  equations  (4)  in  the    unknowns  x,  y,  z.    Five  cases 
arise: 

(a)  D  of  rank  3,  i.e.,  LMO. 

(/3)  D  of  rank  2  {i.e.,  Z>  =  0,  but  some  two-rowed  minor  ^0),  and 


Kv 


hi  bi  c\ 
k2  b2  oi 
h     b3      c3 


K2 


di        k\        C\ 

a2     k2     Ci 
a3     k3     c3 


K3  = 


ai     6i      ki 

Oj        &2         fe 
G&3        &3         &3 


a^ 

h 

bi 

fCf 

Ci 

h 

Oj 

kj 

> 

bj 

kj 

» 

cj 

kj 

not  all  zero. 

(7)  D  of  rank  2  and  Kh  K2,  K3  all  zero. 

(5)  D  of  rank  1  (i.e.,  every  two-rowed  minor  =  0,  but  some  element  ?*0),  and 


(i,  j  chosen  from  1,  2,  3) 


not  all  zero;  there  are  nine  such  determinants  K. 

(e)  D  of  rank  1,  and  all  nine  of  the  two-rowed  determinants  K  zero. 

In  case  (a)  the  equations  have  a  single  set  of  solutions  (§94).  In  cases  (/3)  and 
(5)  there  is  no  set  of  solutions.  For  (/3)  the  proof  follows  from  (13).  In  case  (7)  one 
of  the  equations  is  a  linear  combination  of  the  other  two;  for  example,  if  a\b2  —  a26i^0, 
the  first  two  equations  determine  x  and  y  as  linear  functions  of  z  (as  shown  by  trans- 
posing the  terms  in  z  and  solving  the  resulting  equations  for  x  and  y),  and  the  resulting 
values  of  x  and  y  satisfy  the  third  equation  identically  as  to  z.  Finally,  in  case  (e), 
two  of  the  equations  are  obtained  by  multiplying  the  remaining  one  by  constants. 

The  reader  acquainted  with  the  elements  of  solid  analytic  geometry  will  see  that 
the  planes  represented  by  the  three  equations  have  the  following  relations: 

(a)  The  three  planes  intersect  in  a  single  point. 

()3)  Two  of  the  planes  intersect  in  a  line  parallel  to  the  third  plane. 

(7)  The  three  planes  intersect  in  a  common  line. 

(5)  The  three  planes  are  parallel  and  not  all  coincident. 

(e)   The  three  planes  coincide. 

The  remarks  preceding  our  theorem  furnish  an  illustration  (the  case 
r=n—  1)  of  the  following 

Lemma  1.  If  every  (r+l)-rowed  minor  M  formed  from  certain  r-f-1 
rows  of  D  is  zero,  the  corresponding  r+1  equations  (12)  are  inconsistent 
provided  there  is  a  non-vanishing  determinant  K  formed  from  any  M 
by  replacing  the  elements  of  any  column  by  the  corresponding  known 
terms  kt. 

For  concreteness,1  let  the  rows  in  question  be  the  first  r+1  and  let 

1  All  other  cases  may  be  reduced  to  this  one  by  rearranging  the  n  equations  and 
relabelling  the  unknowns  (replacing  x3  by  the  new  X\,  for  example) 


118 


DETERMINANTS;   SYSTEMS  OF  LINEAR  EQUATIONS    [Ch.VIII 


K 


an 


alr 


&7-+11 


h+lr 


hi 


kT+i 


5^0. 


Let  di,  .  .  .  ,  dr+i  be  the  minors  of  ki,  .  .  .  ,  kr+l  in  K.  Multiply  the 
first  r+1  equations  (12)  by  di,  —d2,  .  .  .  ,  (-l)rdr+i,  respectively,  and 
add.  The  right  member  of  the  resulting  equation  is  the  expansion  of 
±K.     The  coefficient  of  xs  is  the  expansion  of 

an      .  .  .  alr         a,is 


± 


lr+ll 


<T+1  r 


a 


r+ls 


and  is  zero,  being  an  M  if  s>r,  and  having  two  columns  identical  if  s£r. 
Hence  0=  ±K.     Thus  if  K^O,  the  equations  are  inconsistent. 

Lemma  2.  If  all  of  the  determinants  M  and  K  in  Lemma  1  are  zero, 
but  an  r-rowed  minor  of  an  M  is  not  zero,  one  of  the  corresponding  r+ 1 
equations  is  a  linear  combination  of  the  remaining  r  equations. 

As  before  let  the  r+1  rows  in  question  be  the  first  r+1.  Let  the 
non-vanishing  r-rowed  minor  be 


(14) 


dr+l  — 


an 


air 


9*0. 


arl  . 


Let  the  functions  obtained  by  transposing  the  terms  kt  in  (12)  be 

Li  =  anxi+ai2X2Jr  .  .  .  +ainxn  —  kt. 

By  the  multiplication  made  in  the  proof  of  Lemma  1, 

diLi-d2L2+  .  .  .  +(-l)rdr+1Lr+1==FK  =  0. 

Hence  Lr+i  is  a  linear  combination  of  Li,  .  .  .  ,  Lr. 

The  first  part  of  the  theorem  is  true  by  Lemma  1.  The  second  part 
is  readily  proved  by  means  of  Lemma  2.  Let  (14)  be  the  non-vanishing 
r-rowed  minor  of  D.  For  s>r,  the  sth  equation  is  a  linear  combination 
of  the  first  r  equations,  and  hence  is  satisfied  by  any  set  of  solutions  of 
the  latter.  In  the  latter  transpose  the  terms  involving  xT+1,  .  .  .  ,  xn. 
Since  the  determinant  of  the  coefficients  of  xi,  .  .  .  ,  xT  is  not  zero,  §  94 
shows  that  xi,  .  .  .  ,  xT  are  uniquely  determined  linear  functions  of  xT+1, 
.  .  .  ,  xn  (which  enter  from  the  new  right  members). 


97] 


HOMOGENEOUS  EQUATIONS 


119 


EXERCISES 

Apply  the  theorem  to  the  following  four  systems  of  equations  and  check  the  con- 
clusions: 


1.  2x+  y+3z  =  l, 
4x+2y-  z=-3, 

2x+  y—^z  =  —4. 

3.  x-  3y+  4z  =  l, 
4x-12t/  +  162  =  3, 
3x-  9y  +  12z  =  3. 


2.  2x+  y+Sz  =  l, 

4:x+2y-  2  =  3, 
2x+  y  —  42  =  4. 

4.  x—  3y4-  42  =  1, 
4x-  12j/  +  I62  =  4, 
3s-  9y  +  12z  =  3. 


5  Discuss  the  system 


ax+  7/+  z  =  a  — 3, 

x+ay+  z=  —2, 
x+  y+az=-2, 

when    (i)  a  =  l;    (u)  o=— 2;    {Hi)  a^l,— 2,  obtaining  the  simplest    forms    of    the 
unknowns. 

6.  Discuss  the  system 

x+  y+  2  =  1, 
ax+  by-\-  cz  =  k, 
a2x-\-b2y-\-c2z  =  k2, 

when   (i)  a,  b,  c  are  distinct;    (u)  a  =  b^c;    (m)  a  =  b  =  c. 

97.  Homogeneous  Linear  Equations.  When  the  known  terms  fci,  .  .  .  , 
kn  in  (12)  are  all  zero,  the  equations  are  called  homogeneous.  The  determi- 
nants K  are  now  all  zero,  so  that  the  n  homogeneous  equations  are  never 
inconsistent.  This  is  also  evident  from  the  fact  that  they  have  the  set 
of  solutions  £i  =  0,  .  .  .  ,  xn  =  0.  By  (13),  there  is  no  further  set  of  solu- 
tions if  ZMO.  If  D  =  0,  there  are  further  sets  of  solutions.  This  is 
shown  by  the  theorem  of  §  96  which  now  takes  the  following  simpler  form. 

//  the  determinant  D  of  the  coefficients  of  n  linear  homogeneous  equations 
in  n  unknowns  is  of  rank  r,  r<n,  the  r  equations  involving  the  elements  of  a 
non-vanishing  r-rowed  minor  of  D  determine  uniquely  r  of  the  unknowns  as 
linear  functions  of  the  remaining  n  —  r  unknowns,  which  are  independent  vari- 
ables, and  the  expressions  for  these  r  unknowns  satisfy  also  the  remaining 
n  —  r  equations. 

The  particular  case  mentioned  is  the  much  used  theorem: 

A  necessary  and  sufficient  condition  that  n  linear  homogeneous  equations 
in  n  unknowns  shall  have  a  set  of  solutions,  other  than  the  trivial  one  in  which 
each  unknown  is  zero,  is  that  the  determinant  of  the  coefficients  be  zero. 


120  DETERMINANTS;  SYSTEMS  OF  LINEAR  EQUATIONS     [Ch.VIII 

EXERCISES 

Discuss  the  following  systems  of  equations : 

1.  x  +  y+Sz  =  0,  2.  2x-     y+  4z  =  0,  3.     x-  By+  4.2  =  0, 

x+2y+2z=0,  x+  Zy-  2z  =  0,  4x-\2y  +  16z  =  0, 

x+5y-  2=0.  x-lly  +  Uz  =  0.  dx-  9y  +  12z  =  0. 

4.  6x4-4?/4-3z-84tt>  =  0,  5.  2x4-  Zy-  4z  +  5w  =  0, 

x+2y+3z-48w  =  0,  3x+  5y-     z  +  2m;  =  0, 

x-2y+  z-12w  =  0,  7x+lly-  9z+12w  =  0, 

4a; +42/-  z-24w  =  0.  3x  +  4y-llz+13u>  =  0. 

98.  System  of  m  Linear  Equations  in  n  Unknowns.  The  case  m<n 
may  be  treated  by  means  of  the  lemmas  in  §  96.  If  m>n,  we  select 
any  n  of  the  equations  and  apply  to  them  the  theorems  of  §§  94,  96.  If 
they  are  found  to  be  inconsistent,  the  entire  system  is  evidently  inconsist- 
ent. But  if  the  n  equations  are  consistent,  and  if  r  is  the  rank  of  the 
determinant  of  their  coefficients,  we  obtain  r  of  the  unknowns  expressed 
as  linear  functions  of  the  remaining  n—r  unknowns.  Substituting  these 
values  of  these  r  unknowns  in  the  remaining  equations,  we  obtain  a 
system  of  m—  n  linear  equations  in  n  —  r  unknowns.  Treating  this  sys- 
tem in  the  same  manner,  we  ultimately  either  find  that  the  proposed 
m  equations  are  consistent  and  obtain  the  general  set  of  solutions, 
or  find  that  they  are  inconsistent.  To  decide  in  advance  whether  the 
former  or  latter  of  these  cases  will  arise,  we  have  only  to  find  the  maxi- 
mum order  r  of  a  non-vanishing  r-rowed  determinant  formed  from  the 
coefficients  of  the  unknowns,  taken  in  the  regular  order  in  which  they 
occur  in  the  equations,  and  ascertain  whether  or  not  the  corresponding 
(r+l)-rowed  determinants  K,  formed  as  in  §  96,  are  all  zero. 

The  last  result  may  be  expressed  simply  by  employing  the  terminology 
of  matrices.  The  system  of  coefficients  of  the  unknowns  in  any  set  of 
linear  equations 

an  a;i+  •  •  •    -+alnxn=ki, 
(15)  

a^jl  -El    I      •   •   •         \0'mn  %n       >^mj 

arranged  as  they  occur  in  the  equations,  is  called  the  matrix  of  the  coeffi- 
cients, and  is  denoted  by 

/  an   ai2  .  .  .  alw   \ 
A= 


§98]  m  EQUATIONS  IN  n  UNKNOWNS;  MATRIX  121 

By  annexing  the  column  composed   of  the  known  terms  kt,  we  obtain 
the  so-called  augmented  matrix 

I  an  ai2  .  .  .  ain  k\   \ 

*- • 

\    Q'm\    Q"m2  •   •    •    "mn    K>m    I 

The  definitions  of  an  r-rowed  minor  (determinant)  of  a  matrix  and  of 
the  rank  of  a  matrix  are  entirely  analogous  to  the  definitions  in  §  95. 

In  view  of  Lemma  1  in  §  96,  our  equations  (15)  are  inconsistent  if 
B  is  of  rank  r+1  and  A  is  of  rank£r.  By  Lemma  2,  if  A  and  B  are  both 
of  rank  r,  all  of  our  equations  are  linear  combinations  of  r  of  them.  Noting 
also  that  the  rank  r  of  A  cannot  exceed  the  rank  of  B,  since  every  minor 
of  A  is  a  minor  of  B,  and  hence  a  non-vanishing  r-rowed  minor  of  A  is  a 
minor  of  B,  so  that  the  rank  of  B  is  not  less  than  r,  we  have  the  following 

Theorem.  A  system  of  m  linear  equations  in  n  unknowns  is  consistent 
if  and  only  if  the  rank  of  the  matrix  of  the  coefficients  of  the  unknowns  is 
equal  to  the  rank  of  the  augmented  matrix.  If  the  rank  of  both  matrices  is  r, 
certain  r  of  the  equations  determine  uniquely  r  of  the  unknowns  as  linear 
functions  of  the  remaining  n  —  r  unknowns,  which  are  independent  vari- 
ables, and  the  expressions  for  these  r  unknowns  satisfy  also  the  remaining 
m  —  r  equations. 

When  m  =  n-\-l,  B  has  an  m-rowed  minor  called  the  determinant  of 
the  square  matrix  B.  If  this  determinant  is  not  zero,  B  is  of  rank  m. 
Since  A  has  no  m-rowed  minor,  its  rank  is  less  than  m.  Hence  we  obtain 
the 

Corollary.  Any  system  of  n-\-l  linear  equations  in  n  unknowns  is 
inconsistent  if  the  determinant  of  the  augmented  matrix  is  not  zero. 

EXERCISES 

Discuss  the  following  systems  of  equations: 

1.     2x+  y+3z=l,  2.  2x-  y+Bz  =  2,  3.  4x-  y+  z  =  5,  4.     4x-5y  =  2, 

4x+2y-  z= -3,  x+7y+  s  =  l,  2x-3y+5z  =  l,               2x+3y  =  12, 

2x+  y-\z=-\,  3x+5y-5z  =  a,  x+  y-2z  =  2,             10x-7y  =  lQ. 

10x+5t/-6z=-10.  4x-By+2z  =  l.  5x         -  z  =  2. 

5.  Prove  the  Corollary  by  multiplying  the  known  terms  by  xn+i  =  l  and  applying 
§  97  with  n  replaced  by  n  +  1. 

6.  Prove  that  if  the  matrix  of  the  coefficients  of  any  system  of  linear  homogeneous 
equations  in  n  unknowns  is  of  rank  r.  the  values  of  certain  n— r  of  the  unknowns  mav  be 


122 


DETERMINANTS;  SYSTEMS  OF  LINEAR  EQUATIONS     [Ch.VIII 


assigned  at  pleasure  and  the  others  will  then  be  uniquely  determined  and  satisfy  all 
of  the  equations. 

99.  Complementary  Minors.     The  determinant 

a\     61     c\     d\ 

(16)  D  = 


0,2       &2       C2       (1% 

a%     63     C3     ds 
a±     &4     C4    d± 


is  said  to  have  the  two-rowed  complementary  minors 


M  = 


a$  63 


M'  = 


C2  d2 

C4  d\ 


since  either  is  obtained  by  erasing  from  D  all  the  rows  and  columns  having 
an  element  which  occurs  in  the  other. 

In  general,  if  we  erase  from  a  determinant  D  of  order  n  all  but  r  rows 
and  all  but  r  columns,  we  obtain  a  determinant  M  of  order  r  called  an 
r-rowed  minor  of  D.  But  if  we  had  erased  from  D  the  r  rows  and  r  columns 
previously  kept,  we  would  have  obtained  an  (n  —  r) -rowed  minor  of  D 
called  the  minor  complementary  to  M.  In  particular,  any  element  is 
regarded  as  a  one-rowed  minor  and  is  complementary  to  its  minor  (of 
order  n—  1). 

100.  Laplace's  Development  by  Columns.  Any  determinant  D  is 
equal  to  the  sum  of  all  the  signed  products  ±MM',  where  M  is  an  r-rowed 
minor  having  its  elements  in  the  first  r  columns  of  D,  and  M'  is  the  minor 
complementary  to  M,  while  the  sign  is  +  or  —  according  as  an  even  or  odd 
number  of  interchanges  of  rows  of  D  will  bring  M  into  the  position  occupied 
by  the  minor  M\  whose  elements  lie  in  the  first  r  rows  and  first  r  columns 
ofD. 

For  r  =  l,  this  development  becomes  the  known  expansion  of  D  according  to  the 

elements  of  the  first  column  (§  90);  here  Mi  =  en. 
If  r  =  2  and  D  is  the  determinant  (16), 


D  = 

a2   b2 

Cz    dz 
C4    ci\ 

- 

ai    bi 
a3    b3 

c2    d2 
Ci    di 

+ 

ai    bi 
ai    bi 

Ci    di 

cz    di 

+ 

Oi    b2 
a3   h 

' 

ci    di 
Ci    di 

- 

G&2     bi 

04    bi 

Ci    di 
cz     dz 

+ 

03    63 
ai   bi 

Ci    di 
c2    di   ] 

§100] 


LAPLACE'S  DEVELOPMENT 


123 


The  first  product  in  the  development  is  M1M1';  the  second  product  is  —MM'  (in  the 
notations  of  §  99) ,  and  the  sign  is  minus  since  the  interchange  of  the  second  and  third 
rows  of  D  brings  this  M  into  the  position  of  Mi.  The  sign  of  the  third  product  in 
the  development  is  plus  since  two  interchanges  of  rows  of  D  bring  the  first  factor 
into  the  position  of  Mi. 


If  D  is  the  determinant  (8) ,  then 


Mr 


en 


Mi'  = 


■r  +  1  r+l 


er+ln 


&n  t+1 


Any  term  of  the  product  M \M \  is  of  the  type 
(17)  (-l)Vi  ei22  .  .  .  eirT-  (-l)\+ir+i  .  .  .  ev, 

where  i\,  .  .  .  ,  iT  is  an  arrangement  of  1,  .  .  .  ,  r  derived  from  1,  .  .  .  ,  r 
by  i  interchanges,  while  ir+1,  .  .  .  ,  in  is  an  arrangement  of  r+l,  .  .  .  ,  n 
derived  by  j  interchanges.  Hence  i\,  .  .  .  ,  in  is  an  arrangement  of 
1,  .  .  .  ,  n  derived  by  i-\-j  interchanges,  so  that  the  product  (17)  is  a  term 
of  D  with  the  proper  sign. 

It  now  follows  from  §  87  that  any  term  of  any  of  the  products  ±MM' 
mentioned  in  the  theorem  is  a  term  of  D.  Clearly  we  do  not  obtain  twice 
in  this  manner  the  same  term  of  D. 

Conversely,  any  term  t  of  D  occurs  in  one  of  the  products  ±MM'. 
Indeed,  t  contains  as  factors  r  elements  from  the  first  r  columns  of  D, 
no  two  being  in  the  same  row,  and  the  product  of  these  is,  except  per- 
haps as  to  sign,  a  term  of  some  minor  M.  Thus  t  is  a  term  of  MM'  or 
of  —MM'.  In  view  of  the  earlier  discussion,  the  sign  of  t  is  that  of  the 
corresponding  term  in  ±MM',  where  the  latter  sign  is  given  by  the 
theorem. 

101.  Laplace's  Development  by  Rows.  There  is  a  Laplace  develop- 
ment of  D  in  which  the  r-rowed  minors  M  have  their  elements  in  the  first 
r  rows  of  D,  instead  of  in  the  first  r  columns  as  in  §  100.  To  prove  this, 
we  have  only  to  apply  §  100  to  the  equal  determinant  obtained  by  inter  • 
changing  the  rows  and  columns  of  D. 

There  are  more  general  (but  less  used)  Laplace  developments  in  which 
the  r-rowed  minors  M  have  their  elements  in  any  chosen  r  columns  (or 
rows)  of  D.  It  is  simpler  to  apply  the  earlier  developments  to  the  determi- 
nant zfc-D  having  the  elements  of  the  chosen  r  columns  (or  rows)  in  the 
new  first  r  columns  (or  rows) . 


124  DETERMINANTS;  SYSTEMS  OF  LINEAR  EQUATIONS     [Ch.  VIII 

EXERCISES 

1.  Prove  that 


abed 

e    f     g     h 

a     b 

j     k 

0    0    j     k 

e    f 

I    m 

0    0    I    m 

2.  By  employing  2-rowed  minors  from  the  first  two  rows,  show  that 


abed 

e    f     g     h 
abed 

= 

a  b 
e  f 

c  d 

g  h 

1  a  c 

e  g 

b  d 
f  h 

+ 

e    f     g     h 

a  d 
e  h 


b  c 
f  9 


=  0. 


3.  By  employing  2-rowed  minors  from  the  first  two  columns  of  the  4-rowed  deter- 
minant in  Ex.  2,  show  that  the  products  in  Laplace's  development  cancel. 

102.  Product  of  Determinants.  The  product  of  two  determinants  of 
the  same  order  is  equal  to  a  determinant  of  like  order  in  which  the  element 
of  the  r  th  row  and  c  th  column  is  the  sum  of  the  products  of  the  elements  of 
the  r  th  row  of  the  first  determinant  by  the  corresponding  elements  of  the  c  th 
column  of  the  second  determinant. 


For  example, 


(18) 


a     b 

e    f 

c     d 

g    h 

ae-\-bg 
ce+dg 


af+bh 
cf+dh 


While  for  brevity  we  shall  give  the  proof  for  determinants  of  order  3, 
the  method  is  seen  to  apply  to  determinants  of  any  order.  By  Laplace's 
development  with  r  =  3  (§  101),  we  have 


(19) 


a\ 

bi 

C\ 

0    0 

0 

tt2 

b2 

C-2 

0    0 

0 

az 

bs 

Oi 

0    0 

0 

-1 

0 

0 

ei  fi 

9i 

0  - 

-1 

0 

62  J2 

92 

0 

0   - 

-1 

63  h 

93 

a\  &i  ci 

ei  h 

01 

a2    b%    C2 

&2  J2 

92 

a3  63  C3 

63  /3 

93 

102] 


PRODUCT  OF  DETERMINANTS 


125 


In  the  determinant  of  order  6,  add  to  the  elements  of  the  fourth,  fifth, 
and  sixth  columns  the  products  of  the  elements  of  the  first  column  by 
ei,  /i,  gi,  respectively  (and  hence  introduce  zeros  in  place  of  the  former 
elements  ei,  /i,  g{).  Next,  add  to  the  elements  of  the  fourth,  fifth,  and 
sixth  columns  the  products  of  the  elements  of  the  second  column  by 

62,  hi  92,  respectively.     Finally,  add  to  the  elements  of  the  fourth,  fifth, 
and  sixth  columns  the  products  of  the  elements  of  the  third  column  by 

63,  /3,  #3,  respectively.     The  new  determinant  is 


aiei+&i?2+cie3  ai/i+&i/2+ei/3  ai0i+&igr2  +  ci03 

^261  +  6262  +  0263  «2/l+&2/2  +  C2/3  a2#l  +  &202  +  C203 

a3ei+63e2+C3e3  03/1+63/2  +  63/3  03^1 Jrb3g2 +63^3 
0                            0  0 

0  0  0 

0  0  0 


ai 

61 

61 

Ct2 

62 

62 

Ct3 

&3 

63 

1 

0 

0 

0 

-1 

0 

0 

0 

-1 

By  Laplace's  development  (or  by  expansion  according  to  the  elements  of 
the  last  row,  etc.),  this  is  equal  to  the  3-rowed  minor  whose  elements 
are  the  long  sums.  Hence  this  minor  is  equal  to  the  product  in  the  right 
member  of  (19). 


EXERCISES 


1.  Prove  (18)  by  means  of  §  92. 

2.  Prove  that,  if  8i=ai+pt+yi, 


111 

a       /3       7 
a2     /32     72 


1 

a 

a2 

3 

Si 

S2 

1 

0 

P 

= 

Si 

S2 

S3 

1 

Y 

72 

s2 

S3 

Si 

3.  If  Ai,  Bi,  Ci  are  the  minors  of  a*,  bi,  c%  in  the  determinant  D  defined  by  the  second 
factor  below,  prove  that 


Ai 

-A, 

A, 

a\    bi    Ci 

D    0     0 

-Bi 

B, 

-Bz 

a-i    b-i    c-i 

= 

0    D     0 

& 

-c2 

C3 

a%    b3    ^ 

0     0    D 

Hence  the  first  factor  is  equal  to  D2  if  D  ?^0. 


126 


DETERMINANTS;  SYSTEMS  OF  LINEAR  EQUATIONS     [Ch.VIII 


4.  Express  (a2 +b2 +c2 -\-d2) (e2  +/2 -\-g2 -\-h2)  as  a  sum  of  four  squares  by  writing 


a-\-bi      c-\-di 
-c-\-di     a  —  bi 


e+fi      g+hi 
■g+hi      e—fi 


as  a  determinant  of  order  2  similar  to  each  factor.     Hint:   If  k'  denotes  the  conjugate 
of  the  complex  number  k,  each  of  the  three  determinants  is  of  the  form 

k     I 
-V   k' 


MISCELLANEOUS  EXERCISES 

1.  Solve 

ax-\-  by-\-  cz  =  k, 

a2x-\-b2y-\-c2z  =  k2, 
a4x+biy+ciz  =  ki 

by  determinants  for  x,  treating  all  cases. 

2.  In  three  linear  homogeneous  equations  in  four  unknowns,  prove  that  the  values 
of  the  unknowns  are  proportional  to  four  determinants  of  order  3  formed  from  the 
coefficients. 

Factor  the  following  determinants: 


a     be 
b     ca 


ab 
c 
b 
a 


4. 


X 

x2  yz 

X2     X3     1 

y 

y2  xz 

= 

y2  y3   l 

z 

z2    xy 

z2    z3    1 

=  (a+6+c)(a+&G)+c«2)(a+frw2+c«), 


where  a>  is  an  imaginary  cube  root  of  unity. 


(). 


a 

b 

c 

d 

b 

a 

d 

c 

c 

d 

a 

b 

d 

c 

b 

a 

7. 


a  b 

d  a 

c  d 

b  c 


8.  If  the  points  (xi,  i/i), 


d    a 

.  ,  (x4,  y^)  lie  on  a  circle,  prove  that 
xi2+yi2         xi         ?/i  1 


X42  +  2/42  Xi 


Vi 


{). 


§102] 


MISCELLANEOUS  EXERCISES 


127 


9.  Prove  that 


aa'  -\-W  +cc' 
ae'+bf'+cg' 


+ 


ea'+fb'+gc' 

ee'+ff'+gg' 
b 


+ 


f    g 


10.  Prove  that  the  cubic  equation 


a—x        b           c 

D(x)  = 

b       f-x        g 
c           g       h—x 

=  0 

has  only  real  roots.     Hints: 

a2+b2+  c2  — x2      ab+bf+  eg             ac+bg-\- eh 

D(x)-D(-x)  = 

ab+bf  +cg             b2  +  f2+  g2-x2     be  +fg+gh 

ac-\-bg-\-c 

h             be  -\-fg-\-gh 

c2  +g2  +  h2-x* 

=  -x*+xi(a2+P+h2+2b2+2c2+2g2)-x2(D1+D2+D3)+D2(0), 

where  Dz  denotes  the  first  determinant  in  Ex.  9  with  all  accents  removed  and  with 
e  =  b,  while  A  and  Z)2  are  analogous  minors  of  elements  in  the  main  diagonal  of  the 
present  determinant  of  order  3  with  x  =  0.  Hence  the  coefficient  of  —  x2  is  a  sum  of 
squares.  Since  the  function  of  degree  6  is  not  zero  for  a  negative  value  of  x2,  D(x)  =0 
has  no  purely  imaginary  root.  If  it  had  an  imaginary  root  r+si,  then  D(x+r)  =0 
would  have  a  purely  imaginary  root  si.  But  D(x+r)  is  of  the  form  D(x)  with  a,  f,  h 
replaced  by  a—r,  f—r,  h—r.  Hence  D(x)  =0  has  only  real  roots.  The  method  is 
applicable  to  such  determinants  of  order  n. 

11.  If  oi,  .  .  .  ,  an  are  distinct,  solve  the  system  of  equations 


Xi x2 

:  —  a-i     h 


<H 


■  + 


+ 


xn 


h 


=  1 


(t  =  l,  .  .  .  ,  n). 


Hint:  Regard  ki,  .  .  .  ,  kn  as  the  roots  of  an  equation  of  degree  n  in  k  formed  from 
the  typical  one  above  by  substituting  k  for  ki  and  clearing  of  fractions;  write  k  =  aj—t, 
and  consider  the  product  of  the  roots  of  tn-\-  ...  =0.     Hence  find  xj. 

12.  Solve  the  equation 

a+x       x 


b+x 
x 


X 
X 

c-\-x 


=  0. 


CHAPTER  IX 

Symmetric  Functions 

103.  Sigma  Functions,  Elementary  Symmetric  Functions.  A  rational 
function  of  the  independent  variables  x\,  X2,  .  .  .,  xn  is  said  to  be  symmetric 
in  them  if  it  is  unaltered  by  the  interchange  of  any  two  of  the  variables. 
For  example, 

Xl2 +X22  +  XS2 +4^1  +4Z2 +4£3 

is  a  symmetric  polynomial  in  xi,  X2,  £3;  the  sum  of  the  first  three  terms 
is  denoted  by  2zi2  and  the  sum  of  the  last  three  by  42zi.  In  general, 
if  t  is  a  rational  function  of  x\,  .  .  .  ,  xn,  22  denotes  the  sum  of  t  and  all 
of  the  distinct  functions  obtained  from  t  by  permutations  of  the  variables ; 
such  a  2-function  (read  sigma  function)  is  symmetric  in  xi,  .  .  .  ,  xn. 
For  example,  if  there  are  three  independent  variables  a,  fi,  7, 

Za(3  =  al3+ay+(3y,  Ha2p  =  a2P+ap2+a2y+ay2+p2y+py2, 

^a     a     p     7  ^^a.     a     p     7     p     7     a 

^a2+/32     a2+/32  ,  a2+72  .  /32+72 


2! 


a/3  «/3  0:7  j3y 


In  particular,  2a=a+/3+7,  2a/3,  and  a/37  are  called  the  three  elementary 
symmetric  functions  of  a,  /3,  7.     In  general, 

2a  1,  2aia2,         2aia2a3,  .  .  .  ,  2aia2  .  .  .  an_i,  aia2  .  .  .  a„ 

are  the  elementary  symmetric  functions  of  a\,  012,  .  .  ■  ,  an.  In  §  20  they 
were  written  out  more  fully  and  proved  to  be  equal  to  —c\,  C2,  —  cs,  .  .  .  , 
(  —  l)ncn  if  ai,  .  .  .  ,  an  are  the  roots  of  the  equation 

(1)  xn+c1xn-1+c2xn-2+  .  .  .  +c„  =  0 

whose  leading  coefficient  is  unity. 

128 


§  104]        FUNDAMENTAL  THEOREM  ON  SYMMETRIC  FUNCTIONS      129 

EXERCISES 

If  a,  /3,  7  are  the  roots  of  xz-{-px2-{-qx-{-r  =  0,  so  that  2a  =  —p,  2a/3  =  q,  and  a/3-y  =  —  r, 
prove  that 

1.  (2a)2  =  2a2+22a/3,  whence  2a2  =  p2-2g. 

2 .  2a  •  2a/3  =  2a2/3 +3a/37,  whence  2a20  =  3r  -  pg . 

3.  2a2/?7=p7\ 

4.  2a2|32  =  (2a/3)2-2a/372a=g2-2pr. 

If  a,  0,  7,  5  are  the  roots  of  xi+px3+qx2+rx-\-s  =  0,  prove  that 
■sr-M     —  r  ''SAJL_?  V  1     r2_2<7s 

^-^ a.        S   '  ^a^      s'  -*— •  a2  s2 

Hint:   Compute  the  sum,  sum  of  the  products  two  at  a  time,  and  sum  of  the  squares 
of  the  roots  of  the  equation 

l+py+qy2+ry3+syi  =  0, 

obtained  by  replacing  x  by  1/y  in  the  given  quartic  equation. 

iV»_y,yi_4_a:-«,         7.  y^=y?. 

>Wa      ^— ^       X^a  s  -^W     a/3  ^a 


g     V-=  Va/3-  V--V^ 
^-Va2      ^w"P    ^Wa2       ^a 


0      1  / 

— (qr2  —  2q2s — prs +4s2) . 
s2 


a     ^Q  7  _3r-pg 
^Wa/3  s 

104.  Fundamental  Theorem  on  Symmetric  Functions.  Am/  pofo/- 
nomial  symmetric  in  x\,  .  .  .  ,  xn  is  equal  to  an  integral  rational  function, 
with  integral  coefficients,  of  the  elementary  symmetric  functions 

(2)     Ei  =  Xxh         E2  =  2xix2,         E3  =  '2xiX2X3,  .  .  .  ,        En  =  x1x2  .  .  .  xn 

and  the  coefficients  of  the  given  polynomial.  In  particular,  any  symmetric 
polynomial  with  integral  coefficients  is  equal  to  a  polynomial  in  the  elementary 
symmetric  functions  with  integral  coefficients. 

For  example,  if  n  =  2, 

rx12+rx22+sxl+sx2=r(El2-2E2)+sEl. 

In  case  r  and  s  are  integers,  the  resulting  polynomial  in  E\  and  E2  has  integral  coeffi- 
cients. 

The  theorem  is  most  frequently  used  in  the  equivalent  form: 


130  SYMMETRIC  FUNCTIONS  [Ch.  IX 

Any  polynomial  symmetric  in  the  roots  of  an  equation, 

xn-E1xn-1+E2xn-2-  .  .  .  +  (-l)nEn  =  0, 

is  equal  to  an  integral  rational  function,  with  integral  coefficients,  of  the  coeffi- 
cients of  the  equation  and  the  coefficients  of  the  polynomial. 

It  is  this  precise  theorem  that  is  required  in  all  parts  of  modern  algebra 
and  the  theory  of  numbers,  where  attention  to  the  nature  of  the  coefficients 
is  vital,  rather  than  the  inadequate,  oft-quoted,  theorem  that  any  sym- 
metric function  of  the  roots  is  expressible  (rationally)  in  terms  of  the 
coefficients. 

It  suffices  to  prove  the  theorem  for  any  homogeneous  symmetric  poly- 
nomial S,  i.e.,  one  expressible  as  a  sum  of  terms 

h  =  ax\^X2ki  •  .  •  Xn*" 

of  constant  total  degree  k  =  ki+k2+  •  •  •  +kn  in  the  x's.  Evidently 
we  may  assume  that  no  two  terms  of  S  have  the  same  set  of  exponents 
fci,  .  .  .  ,  kn  (since  such  terms  may  be  combined  into  a  single  one).  We 
shall  say  that  h  is  higher  than  the  term  bx\hX2h  .  .  .  xnln  if  k\>l\,  or  if 
ki  =  h,  k2>h,  or  if  ki  =  h,  k2  =  h,  kz>h,  .  .  .  ,  so  that  the  first  one  of 
the  differences  k\  —  l\,  k2  —  h,  k^—h,  .  .  .  which  is  not  zero  is  positive. 

We  first  prove  that,  if  the  above  term  h  is  the  highest  term  of  S,  then 
k\~t.k2^kz  .  .  .  ^kn. 
For,  if  ki<k2,  the  symmetric  polynomial  S  would  contain  the  term 

ax\kiX2klxz^  .  .  .  xnkn, 
which  is  higher  than  h.     If  /c2<&3,  S  would  contain  the  term 

aXikiX2k3X3k2  .  .  .  Xn**, 

which  is  higher  than  h,  etc. 

If  the  highest  term  in  another  homogeneous  symmetric  polynomial 
S'is 

h'  =  a'xiki'x2k2'  .  .  .  xnkn', 

and  that  of  S  is  h,  then  the  highest  term  in  their  product  SS'  is 
hh'  =  aa'xikl+!ci  .  .  .  xntn+kn'. 


§  104]      FUNDAMENTAL  THEOREM  ON  SYMMETRIC  FUNCTIONS        131 

Indeed,  suppose  that  SS'  has  a  term,  higher  than  hh', 

(3)  cxi'i+'i'  .  .  .  xnln+ln', 

which  is  either  a  product  of  terms 

t  =  bx\K  .  .  xnln,         t'  =  b'xili'  .  .  .  XrS 

of  S  and  S'  respectively,  or  is  a  sum  of  such  products.  Since  (3)  is  higher 
than  hh',  the  first  one  of  the  differences 

Li-Til        k\       K\  ,  .  .  .  ,  ln~\~('n        l^n      ^n 

which  is  not  zero  is  positive.  But,  either  all  of  the  differences  h—ki, 
.  .  .  ,  ln—kn  are  zero  or  the  first  one  which  is  not  zero  is  negative,  since 
h  is  either  identical  with  t  or  is  higher  than  t.  Likewise  for  the  differences 
li'  —  ki,  .  .  .  ,  ln—kn  .     We  therefore  have  a  contradiction. 

It  follows  at  once  that  the  highest  term  in  a  product  of  any  number 
of  homogeneous  symmetric  polynomials  is  the  product  of  their  highest 
terms.     Now  the  highest  terms  in  Ei,  E2,  E3,  .  .  .  ,  En,  given  by  (2),  are 

Xl,  X\X2,  XxXlXz,       .   .   .  ,  X\X2       .   .   .  Xn, 

respectively.     Hence  the  highest  term  in  Eia^E2a2  .  .  .  Enan  is 

Thus  the  highest  term  in 

<j  =  aEiki-k*E2k2-k3  .  .  .  En_1kn-i-knEnk» 

is  h.  Hence  Si=S  —  <r  is  a  homogeneous  symmetric  polynomial  of  the 
same  total  degree  k  as  S  and  having  a  highest  term  hi  not  as  high  as  h. 
As  before,  we  form  a  product  a\  of  the  E's  whose  highest  term  is  this  hi. 
Then  S2  =  Si  —  ai  is  a  homogeneous  symmetric  polynomial  of  total  degree 
k  and  with  a  highest  term  /12  not  as  high  as  hi.  We  must  finally  reach 
a  difference  Si—(Tt  which  is  identically  zero.  Indeed,  there  is  only  a 
finite  number  of  products  of  powers  of  xi,  .  .  .  ,  xn  of  total  degree  k. 
Among  these  are  the  parts  h',  h'i,  h'2,  ...  of  /i,  hi,  /12,  •  •  •  with  the  coeffi- 
cients suppressed.  Since  each  ht  is  not  as  high  as  ht-i,  the  h',  hi,  h2,  .  .  .  are 
all  distinct.      Hence  there  is  only  a  finite  number  of  ht.     Since  St—<rt  =  0, 

S  =  a-\-Si  =  <T-\-ai-\-S2=   .  .  .  =o"  +  o"l  +  o"2+  •  •  •  -\~o-f 

Hence  £  is  a  polynomial  in  Ei,  E2,  .  .  .  ,  En  and  a,  b,  ...  ,  with  integral 
coefficients. 


132  SYMMETRIC  FUNCTIONS  [Ch.  IX 

Example  1.     If  S  =  2xi2x22x3  and  n>  4,  we  have 

a  =  E2E%  =*S+32xi2X2X3X4+10  221X2X3X4X6, 
Si=S— 0=  —3  Sxi2x2x3X4  — 10  2x1X2X3X4X5, 
o\=  —  3  E\Ei=  —3  (2xi2x2x3X4+5  2x1X2X3X4X5), 
Sz  =Si  —  <n  =  5  2x1X2X3X4X5  =  5  .#5, 
*S  =  <r-|-(Si  =  cr-|-(7i4"*Si2  =  E2Ez  — 3  EiEi~{-5  E&. 

Example  2.     If  S  =  2xi3x2x3  and  n>  4, 

o-  =  ^12^3=-Ei(2xi2x2X3+4  2x1X2X3X4) 

=  2xi3x2x3+2  2xi2x22x3+3  2xi2x2X3X4 

+4  (2Xi2X2X3X4  +  5  2X1X2X3X4X5), 

Si=S—<r=  —2  2xi2x22x3— 7  2xi2x2x3X4— 20  2x1X2X3X4X5. 
Take  <n  =  —  2  ^2-^3  and  proceed  as  in  Ex.  1. 

Example  3.     By  examples  1  and  2,  if  n>4, 

o2xi2x22x3+b2xi3x2X3  =  bEJEt  -  (3a +6)^4+ (a  -  2b)E2E3 +5(a +b)E&. 

105.  Rational  Functions  Symmetric  in  all  but  One  of  the  Roots.     If 

P  is  a  rational  function  of  the  roots  of  an  equation  f(x)  =  0  of  degree  n  and 
if  P  is  symmetric  in  n—\  of  the  roots,  then  P  is  equal  to  a  rational  function, 
with  integral  coefficients,  of  the  remaining  root  and  the  coefficients  of  f(x)  and  P. 

For  example,  P  =  rai+a22+a32+  .  .  .  +aw2  is  symmetric  in  a2,  ...  ,  %,  and 

P  =  ra1  +  2o;i2-ai2  =  Ci2-2c2+rai-ai2, 

if  on,  .  .  .  ,  ocn  are  the  roots  of  equation  (1). 

Since x  any  symmetric  rational  function  is  the  quotient  of  two  symmet- 
ric polynomials,  the  above  theorem  will  follow  if  proved  for  the  case  in  which 
the  words  rational  function  are  in  both  places  replaced  by  polynomial. 

If  a\  is  the  remaining  root,  the  polynomial  P  is  symmetric  in  the  roots 
«2,  .  .  .  ,  an  of  f(x)/(x— ai)  =0,  an  equation  of  degree  n—1  whose  coeffi- 
cients are  polynomials  in  a\,  c\,  .  .  .  ,  cn  with  integral  coefficients.  Hence 
(§  104),  P  is  equal  to  a  polynomial,  with  integral  coefficients,  inai,  ci,  .  .  .  , 
cn  and  the  coefficients  of  P. 

1  If  N/D  is  symmetric  in  «i,  «2,  and  the  polynomials  N  and  D  have  no  common 
factor,  while  TV  becomes  N'  and  D  becomes  D'  when  ah  a2  are  interchanged,  then 
ND'  =DN'.  Thus  N  divides  N'  and  both  are  of  the  same  degree.  Hence  N'  =  cN, 
D'=cD,  where  c  is  a  constant.  By  again  interchanging  a\,  a2,  we  obtain  N  from  N', 
whence  N  =  cN'  =  c2N,  c2  =  l.  If  c=—  1,  we  take  ai=a2  and  see  that  N  =  N'=—N, 
N  =  0,  whence  N  has  the  factor  a„—  a2.  Similarly,  D  has  the  same  factor,  contrary  to 
hypothesis.     Hence  c=  +1  and  N  and  D  are  each  symmetric  in  ah  a2. 


§  105]  FUNCTIONS  SYMMETRIC  IN  ALL  BUT  ONE  ROOT  133 

Example.     If  a,  13,  y  are  the  roots  of  f(x)  =xs+px2+qx-\-r  =  0,  find 

^-4  a- 


a!jf_a4f     a2+72     /32+72 
i+P  ~  a+j3        a+y         /3+T  ' 

Solution.     Sincel32-\-y2  =  p2  —  2q—a2,      j3+y=—p—a, 


^a2+(32_^pp2-2q-a2_^p  /  2g\_ yj_ 

But  a+p,  /3+p,  7+p  are  the  roots  yh  y%,  y3  of  the  cubic   equation  obtained  from 
/(x)=0  by  the  substitution  x-\-p  =  y,  i.e.,  x  =  y  —  p.     The  resulting  equation  is 

y3  —  2py2  +  (p2+q)y+r-pq  =  0. 

Since  we  desire  the  sum  of  the  reciprocals  of  yi,  y2,  yz,  we  set  y  =  l/z  and  find  the  sum 
of  the  roots  Z\,  z2,  Zz  of 

1  —2pz-\-  (p2  -\-q)z2  -\-  (r  —  pq)z3  =  0 . 
Hence 

lQ,2+/32     2q2-2p2q+4:pr 


^A      1      _^1_^  p2+g  -ST^o:2 

<^-*a-\-p      ^^ y\      -*-^         pq  —  r  -^-W  a 


+/3  pq—r 


EXERCISES 
[In  Exs.  1-12,  a,  13,  y  are  the  roots  of  f(x)  =  x3+px2+qx+r  =  0.] 
Using  (3y -\-a((3 -\-y)  =  </,  find 


ST^Py+a2  >^p3/37-2a2 


/3+T  '    ^   /3+7-«' 

3.  Why  would  the  use  of  0y  =  —r/a  complicate  Exs.  1,  2?     Verify  that 

— r    f(a)—r 

@y  = = =a2+pa+q. 

a  a 

>V^/32+72 

4.  Why  would  vou  use  fiy=  —r/a  in  finding    /    ? 

*—1  $y  +c 

5.  Find  ^2\{&+y)2.  6.  Find  ^\{a+p-y)\  7.  Find  ^(^^J- 

8.  Find  a  necessary  and  sufficient  condition  on  the  coefficients  that  the  roots,  in 

112  —  3r 

some  order,  shall  be  in  harmonic  progression.     Hint:     If  — | —  =  -,  then — /S  =  0, 

a     7     /3  q 

and  conversely.     Hence  the  condition  is 


134  SYMMETRIC  FUNCTIONS  [Ch.  IX 

1  1  1      TT. 

9.  Find  the  cubic  equation  with  the  roots  /3y  —  —,      ay  —  -,       a/3  —  -.     Hint:   since 

a  /3  y 

these  are  (—  r  —  l)/a,  etc.,  make  the  substitution  (—  r— l)/x=y. 

Find  the  substitution  which  replaces  the  given  cubic  equation  by  one  with  the  roots 

10.  afi+ay,       <x{i  +  fiy,         ay+/3y. 

„       2a-l  ,„      /3T+3a2 

11.  — ,  etc.  12.  — — ,  etc. 

j3+7—  a  /3+y—  <«* 

If  a,  /3,  7,  5  are  the  roots  of  z4+px3+gx2+ra:+s  =  0,  find 

13'    ^   /3+T  +  5  '  '    ^W/3+7  +  5-3" 

15.  Prove  that  if  yh  y2,  ijz  are  the  roots  of  yz+py-\-q  =  0,  the  equation  with  the  roots 
zi  =  (y2-yz)2,     %  =-(2/1-  2/3) 2,     23  =  (2/i-y2)2is 

s3+6pz2+9p2z+4p3+27g2  =  0. 

Hints:  since  Zi  =  ~Eyl2-2y2y3-yi2  =  -2p+2q/yi-yi2,  etc.,  we  set  2=  -2p+2q/y-y2. 
By  the  given  equation,  ?/2+p+g/y  =  0.  Thus  the  desired  substitution  is  2=  —  p+Sq/y, 
y  =  3q/(z+p). 

16.  Hence  find  the  discriminant  of  the  reduced  cubic  equation. 

17.  If  Xi,  .  .  .  ,  xn  are  the  roots  of  f(x)  =0,  show  that 


S; 


1         -/'(c) 


'Xi-C        /(c) 
Hint:  xi— c  =  yi,  .  .  .  ,  xn  —  c  =  yn  are  the  roots  of 

/(c+2/)=/(c)+2//'(c)+?/2(     )+  .  .  .  =0, 
as  shown  by  Taylor's  theorem.     Or  we  may  employ  (5)  for  x  =  c. 

106.  Sums  of  Like  Powers  of  the  Roots.     If  a\,  .  .  .  ,  an  are  the  roots  of 

(1)  f(x)=xn+axn-1+c2xn-2-\-  .  .  .  +cn  =  0, 

we  write  si  for  2a  1,  S2  for  2ai2,  and,  in  general, 

st  =  2aifc  =  ai*+a2A:+  .  .  .  +a„*. 
The  factored  form  of  (1)  is 
(4)  f(x)  =  (x-ai)(x-a2)  .  .  .  (x-an). 


§  106]  SUMS  OF  LIKE  POWERS  OF  THE  ROOTS  135 

The  derivative  f'(x)  of  this  product  is  found  by  multiplying  the  derivative 
(unity)  of  each  factor  by  the  product  of  the  remaining  factors  and  adding 
the  results.     Hence 

f'(x)  =  (x—a2)  .  •  .  (x-an)-\r(x-ai)(x-a3)  .  .  .  (x—an)+  .  .  .  , 

(5)  Ax)s/w+/w +  ...  +  /« 


x—ai     x—a.2  x—an 

If  a  is  any  root  of  (1),  f(a)  =  0  and 

f(x)  _f(x)  -/(«)_ xn-an  .       xn~x-an-x.  x-a 

t"Ci r  •  •  •    i  cw_i 

X  —  a.  x—a  x—a  X  —  a  x—a 

=  xn-l+axn-2+a2xn-z+  .  .  .  +a(xn-2+axn-3+  .  .  .  ) 

+c2(xn-3+  ...)+..., 

(6)       !Q  =  x«-i+(a+Cl)xn-2+(a2+Cia+C2)xn-3+  .  .  . 

X  —  a 

+  (a:fc+Ci^-1  +  C2«fc-2+  .  .  .  +C,_1a  +  cft)xra-t-1+ 

Takings  to  be  ai,  .  .  .  ,  anin  turn,  adding  the  results,  and  applying  (5),  we 
obtain 

f'{x)=nxn~lJr(si-\-nci)xn~2-\-(s2JrCiS\-\-nc2)x' ~*-\-  .  .  . 

+  (sifc+cisfc_1+C2SA;_2+  .  .  .  +ck_1s1+nck)xn-t-1+  .... 

The  derivative  of  (1)  is  found  at  once  by  the  rules  of  calculus  (or  by 
§  56)  to  be 

f'(x)=nxn-1+(n-l)c1xn-2+(:n-2)c2Xn-3+  .  .  .  +(n-k)ckxn-k-1+  .... 
Since  this  expression  is  identical  term  by  term  with  the  preceding,  we  have 

si+ci  =  0,         s2+cisi+2c2  =  0,  .  .  .  , 
(7) 

sifc+CisA;_i+C2Sft_2+  .  .  .  +cjfc_1si+A;Cfc  =  0     (fc£w— 1). 

We  may  therefore  find  in  turn  s\,  S2,  .  .  .  ,  sre_j: 

(8)  Sl=—  Cl,  S2  =  C12  —  2C2,  S3=  —  Ci3  +  3CiC2  — 3  C3,  .  .  .      . 

To  find  sn,  replace  x  in  (1)  by  a\,  .  .  .  ,  an  in  turn  and  add  the  resulting 
equations.     We  get 

(9)  •5n  +  ClS„_1  +  C2.S„_2+    .   .   .    +C„_1Si+WC„  =  0. 


LIBRARY  SEISMOLOGJCAL 
OBSERVATORY 


136  SYMMETRIC  FUNCTIONS  [Ch.  IX 

We  may  combine  (7)  and  (9)  into 

(10)  st+cisfc_1+c2sfc_2+  •  •  •  +cft_isi+/ccfc  =  0     (fc=l,2,  .  .  .  ,  n). 
This  set  of  formulas  (10)  will  be  referred  to  as  Newton's  identities. 

The  student  should  be  able  to  write  them  down  from  memory  and,  when 
writing  them,  should  always  check  the  final  one  (9)  by  deriving  it  as  above. 
To  derive  a  formula  which  shall  enable  us  to  compute  the  sk  for 
k>n,  we  multiply  (1)  by  xk~n,  take  x  =  a\,  ...  ,  x  =  an  in  turn,  and  add 
the  resulting  equations.     We  get 

(11)  sfc+cisfe_1+c2sfc_2+  .  .  .  +cns2fe_„  =  0  (k>n). 

Instead  of  memorizing  this  formula,  it  is  preferable  to  deduce  it  for  the 
particular  equation  for  which  it  is  needed,  thus  avoiding  errors  of  substi- 
tution as  well  as  confusion  with  (10). 

Example.     Find  sj  for  xn  —  1  =  0. 

Solution.  Comparing  our  equation  with  (1),  we  have  Ci=0,  .  .  .  ,  cn_i=0,  cn=  —  1. 
Hence  in  (10)  for  k<n,  each  c  is  zero  and  sk  =  0.  But,  for  k  =  n,  (10)  becomes  sn— n  =  0. 
We  may  check  the  latter  by  substituting  each  root  a\,  .  .  .  ,  an  in  our  given  equation 
and  adding.  Finally,  to  find  si  when  l>n,  multiply  our  equation  by  xl~n.  In  the 
resulting  equation  xl — xl~n  =  0  we  substitute  each  root,  add,  and  obtain  sj  —  sj_re  =  0. 
Hence  from  si  we  obtain  an  equal  s  by  subtracting  n  from  I.  After  repeated  subtrac- 
tions, we  reach  a  value  k  for  which  1  £  k  =  n.  Since  st  =  0  or  n  according  as  k<n  or 
k  =  n,  it  follows  that  s^  =  0  or  n  according  as  I  is  not  or  is  divisible  by  n, 

EXERCISES 

1.  For  a  cubic  equation,  s4  =  Ci4— 4ci2c2+4ciC3+2c22. 

2.  For  an  equation  of  degree  ^  4,  s4  =  Ci4— 4ci2c2+4ciC3+2c22— 4;4. 

3.  Find  s2,  s3,  s4,  s5  for  x2  —  px-\-q  =  0. 

4.  Find  sfc  for  x6 -3=0. 

5.  Find  s2,  s3,  s6,  s7  for  a;5  —  px+g  =  0, 

107.  Waring's  Formula  for  sk  in  Terms  of  the  Coefficients.  While 
we  have  learned  how  to  find  si,  s2,  S3,  ...  in  turn  by  Newton's  identities, 
it  is  occasionally  useful  to  know  an  explicit  expression  for  sk,  where  k 
has  an  arbitrary  value.  The  formula  in  question  is  applied  ordinarily 
only  to  a  quadratic  equation 

x2-\-,pxJrq  =  0. 

Accordingly  we  shall  treat  this  case  in  detail.     If  its  roots  are  a  and  /3, 

then 

x2 + px + q  =  (x — a)  (x  —  /3) . 


§  107]  WARING' S  FORMULA  137 

Replace  x  by  1/y  and  multiply  by  y2.    We  get 
(12)  l+py+qy2^(l-ay)(l-Py). 

Taking  derivatives,  we  have 

p+2qy=-ct(l-Py)-p(l-cty). 
Change  of  signs  and  division  by  the  members  of  (12)  gives 
-p-2qy  _     a      ,      ft 


(13) 


1+py+qy2     I— ay     1-/3?/' 


The  identity  in  Ex.  7,  §  14,  with  n  changed  to  k,  may  be  written  in 
the  form 

(14)  ~ sl+r+rM-..  .  .+  r*-^-— . 

Take  r= ay  and  multiply  the  resulting  terms  by  a;  thus 

„2„,  i  J_«t-,*-iJ__ *L 


=  a+a:2l/+  .  .  .  +ay 


1— a?/  1— al/ 

Similarly, 

1^-*+*+  •  •  •  +**"+££■ 

To  show  that  on  adding,  and  writing  sk  for  ak+^,  we  obtain  (15),  we  need 
the  sum  of  the  final  fractions,  which  by  (12)  is 


<t>y  <$>y 


<$>=ak+\l-Py)+$k+\l-ay). 


(l-ay)(l-Py)     l+py+qy2 
Hence 

(15)  T^-+T^-r  =  s1+s2y+  .  .  .  +skyk-1+..4>y.      2, 

I- ay     1-fiy  y  *  1+py+qy2 

where  the  exact  expression  for  4>  is  immaterial. 

Next,  we  seek  an  expansion  of  the  fraction  in  the  left  member  of 
(13).  Its  denominator  will  be  identical  with  that  in  (14)  if  we  choose 
r=  —  py  —  qy2.     Evidently  (14)  may  be  written  in  the  compact  form 

-i  t  - 1  -k 


1  —  r     t  =  0       1—r 
Hence  it  becomes 


.  ,       ,     2=  S  (-iY(py+qy2Y+,  ,  yy,     2, 

1+py+qy2      ,=0  ™     ™         1+py+qy2 


138  SYMMETRIC  FUNCTIONS  [Ch.  IX 

where  ^=(  —  p  —  qy)lc,  although  no  use  will  be  made  of  the  particular 
form  of  the  polynomial  \p.     By  the  binomial  theorem, 

(g+h)) 


(py+qy2Y=  ^~j£r(py)e(<iy2y> 


g\h\ 

where  the  summation  extends  over  all  sets  of  integers  g  and  h,  each  |:0, 
for  which  g+h  =  t,  while  g\  denotes  the  product  of  1.  2,  .  .  .  ,  g  if  ghl, 
but  denotes  unity  if  g  =  0.     Hence 

(16)         i+lylZ2= (y+2^)s(~  1)0+n+lkgW  r3<inys+2h+E> 

E_(-p-^qy)^yk 
i+py+qy2  ' 

where  the  summation  extends  over  all  sets  of  integers  g  and  h,  each  2:0, 
for  which  g-\-h^.k—l. 

Since  the  left  members  of  (15)  and  (16)  are  identically  equal  by  (13), 
their  right  members  must  be  identical,  so  that  the  coefficients  of  y*'1 
in  them  must  be  equal.1  Hence  the  coefficient  sk  of  yk~l  in  (15)  is  equal 
to  the  coefficient  of  yk~1  in  (16),  which  is  made  up  of  two  parts,  correspond- 
ing to  the  two  terms  of  the  factor  p-\-2qy.  When  we  use  the  constant 
term  p,  we  must  employ  from  2  in  (16)  the  terms  in  which  the  exponent 
of  y  is  equal  to  k—1.  But  when  we  use  the  other  term  2qy,  we  must 
employ  from  2  the  terms  in  which  the  exponent  of  y  is  equal  to  k  —  2,  in 
order  to  obtain  the  combined  exponent  k—  1  of  y.  Hence  sk  is  equal  to 
the  sum  of  the  following  two  parts: 

p2(-l)"+A+1  ^pW         (g+2h  =  k-l), 

2g2(-l)g+ft+l(g"|"/t/|)W        (g+2h  =  k-2). 

In  the  upper  sum,  write  i  for  g-\-l,  and  j  for  h.  In  the  lower  sum,  write 
i  for  g,  and  j  for  h-\-l.     Hence 

1  In  fact,  the  (k  —  l)th  derivatives  of  the  two  right  members  are  identical,  and  we 
obtain  the  indicated  result  by  substituting  y  =  0  in  these  two  derivatives  and  equating 
the  results.  Note  that  the  final  terms  in  both  (15)  and  (16)  have  y  as  a  factor  of  their 
(k  —  l)th  derivatives. 


§  107]  WARING'S  FORMULA  139 

where  now  each  summation  extends  over  all  sets  of  integers  i  and  j,  each 
2:  0,  for  which 

(17)  i+2j=k. 

Finally,  we  may  combine  our  two  sums.  Multiply  the  numerator 
and  denominator  of  the  first  fraction  by  i,  and  those  of  the  second  fraction 
by  j.     Thus 


(18)  ft  =  fcZ(-l)«+'^"7    *;w, 


(»+■;•- PL 

W 


since  the  present  fraction  occurred  first  multiplied  by  i  and  second  multi- 
plied by  2j,  and,  by  (17),  the  sum  of  these  multipliers  is  equal  to  k.  Our 
final  result  is  (18),  where  the  summation  extends  over  all  sets  of  integers 
i  and  j,  each  hO,  satisfying  (17). 

If  we  replace  i  by  its  value  k—2j,  and  change  the  sign  of  p,  we  obtain  from  (18) 
the  result  that  the  sum  of  the  k  th  powers  of  the  roots  of  x2—px+q  =  0  is  equal  to 

(19)  ss  =  *f  (-1)^^V-2V 

k     ,    fc-2    i  k(k  —  3)   k_i  2     k(k— 4)(/c— 5)   j._6  3 
=  P~kp        1+^^p        q r^— p        5+.... 

where  K  is  the  largest  integer  not  exceeding  k/2. 

The  product  of  the  roots  is  equal  to  q.  Hence  if  x  denotes  one  root,  the  second 
root  is  q/x.  Thus  Sk  =  x  -\-(q/x)  .  Again,  the  sum  of  the  roots  is  x-\-q/x  =  p.  Regard 
q  as  given  and  p  as  unknown.     Hence,  if  c  is  an  arbitrary  constant,  the  equation 

(20)  p  -kqp       H T~2~q  V        ~  ■  ■  ■  =c 

is  transformed  by  the  substitution  p  =  x+q/x  into 

^&- 

Hence  equation  (20)  may  be  solved  for  p  by  radicals  by  the  method  employed  in  §  43 
for  a  cubic  equation. 


140  SYMMETRIC  FUNCTIONS  [Ch.  IX 

The  above  proof  applies  1  without  essential   change  to  any  equation 
xn+ClXn-i_^_  +  c„  =  0  and  leads  to  the  following  formula  for  the  sum 

of  the  kth  powers  of  its  roots: 

(21)  *  =  kZ(-l)i+  ■  ■  ■  +*  (ri+n!'.'.+rir1)!cin-  *  •  ^ 

where  the  summation  extends  over  all  sets  of  integers  n,  .  .  .  ,  rn,  each 
^0,  for  which  ri+2r2+3r3+  .  .  •  +nrn  =  k.  This  result  (21)  is  known 
as  Waring' s  formula  and  was  published  by  him  in  1762. 

Example.     Letn  =  3,  fc=4.     Then  r,.+2?-2+3r3  =  4  and 

(ri,  r2,  n)  =  (4,  0,  0),         (2,  1,  0),         (1,  0,  1),         (0,  2,  0), 

/3!  2!  1!  1! 

S4  =  4(v^!Cl4-2mCl2C2  +  l!l!ClC3+2!C2: 

=  ci4-4ci2c2+4c1c3+2c22. 

EXERCISES 

1.  For  the  quadratic  z2  —  pz+g  =  0  write  out  the  expressions  for  s2,  s3,  s4,  s5  given  by 
(19),  and  compare  with  those  obtained  from  Newton's  identities  (Ex.  3,  §  106). 

2.  Find  s4  for  a  quartic  equation  by  Waring's  formula. 

3.  For  k  =  5,  (20)  becomes  De  Moivre's  quintic  ps  —  5qp3+5q2p  =  c.  Solve  it  by 
radicals  for  p. 

4.  Solve  (20)  by  radicals  when  k  =  7. 

108.  S-functions  Expressed  in  Terms  of  the  Functions  sk.  Since 
we  have  learned  two  methods  of  expressing  the  sk  in  terms  of  the  coeffi- 
cients, it  is  desirable  to  learn  how  to  express  any  S-polynomial  (and 
hence  any  symmetric  function)  in  terms  of  the  st. 

By  performing  the  indicated  multiplication,  we  find  that 

sasb  =  2aia  •  W  =  2aia+6+m2aiW, 
where  m=l  if  a^b,  m  =  2  if  a  =  b.     Transposing  the  first  term,  which 
is  equal  to  sa+b,  and  dividing  by  m,  we  obtain 

(22)  2aiaa2b  =  -(sasb-sa+b). 

1  See  the  author's  Elementary  Theory  of  Equations,  pp  72-74,  where  there  is  given 
also  a  shorter  proof  by  means  of  infinite  series. 


§  108]  COMPUTATION  OF  SIGMA  FUNCTIONS  141 

In  order  to  compute  Sai4o!23«32  in  terms  of  the  sk,  we  form  the  product 

Sai4  •  Sai3o:22  =  2ai 'o;22  +  2q:i6q;23  +  2ai4a23a:32. 
Making  three  applications  of  (22),  we  get 

S4  (S3S2  —  S5)  =  (S7S2  —  Sg)  +  (s6S3  —  Sg)  +  Sai4a23o:32. 

Hence 

So:i4a23o:32  =  S2S3S4  —  S2S7  —  S3S6  —  S4S5+2S9. 

EXERCISES 

For  a  quartic  equation,  express  in  terms  of  the  si  and  ultimately  in  terms  of  the 
coefficients  Ci,  .  .  .  ,  c4: 

1.   Sai2a22.  2.   Sai3a2. 

3.   '2ai2a.2fXa.  4.  2ai2a22«32. 

5.  If  a^.b  >  c>0,  prove  that 

Sai°a2  a3<:  =  — (saS6Sc  — SaS6+c— S6S04.C— ScSa+6+2s04-{,4.c), 
m 

where  m  —  \  if  a>&,     m=2ifa  =  fe. 

6.  Sai%6a36=i(sas62-saS26— 2s6sa+5+2sa+2&),         a>&>0. 

7.  Sa!%aa3a=-l-(s03-3saS2a+2s3a),  a>0. 

109.  Computation  of  Symmetric  Functions.  The  method  last  explained 
is  practicable  when  a  term  of  the  S-function  involves  only  a  few  distinct 
roots,  the  largeness  of  the  exponents  not  introducing  a  difficulty  in  the 
initial  work  of  expressing  the  2-f unction  in  terms  of  the  sk. 

But  when  a  term  of  the  S-function  involves  a  large  number  of  roots 
with  small  exponents,  we  resort  to  a  method  suggested  by  §  104,  which 
tells  us  which  auxiliary  simpler  symmetric  fuctions  should  be  multiplied 
together  to  produce  our  S-function  along  with  simpler  ones. 

For  example,  to  find  Zxi2x2x3X4,  when  n>4,  we  employ 

E1E4  =  2xi  •  2x1X2X3X4  =  2xi2x2x3X4+5  2x1X2X3X4X5, 

2xi2x2x3X4  =  .Z?ii?4— 5  Eb. 

To  find  2xi2X22x32x4,  employ  ^3^4  =  2x1X2X3-2x1X2X3X4. 

When  many  such  products  of  2-functions  are  to  be  computed,  it  will  save  time 
in  the  long  run  to  learn  and  apply  the  "  method  of  leaders  "  explained  in  the  author's 
Elementary  Theory  of  Equations,  pp.  64-65. 


142  SYMMETRIC  FUNCTIONS  [Ch.  IX 

MISCELLANEOUS  EXERCISES 

Express  in  terms  of  the  coefficients  ci,  .  .  .  ,  cn: 

1.   2ai2a2«3.  2.   2ai2a22«3.  3.   Sai2a22o;3a4.  4.   Sai2a22a32. 

If  a,  /3,  7  are  the  roots  of  x3+px2+qx+r  =  0,  find  a  cubic  equation  with  the  roots 

2  2  2 

5.  a2,  /32,  T2-  6.  a/3,  ay,  0y.  7.    -,  -,  -. 

a    p   7 

8.  a2+/32,     a2+72,     /32+72.  9.  a2+a/3+02,  etc. 

If  a,  /3,  7,  5  are  the  roots  of  x4+px3+qx2+rx+s  =  0,  find 

io.  y^ythrL«.y-z^,_4_pyi. 

jL^ta2'  '     -^Wo:    ^- /a      ^-*a2         -^Wa         ^Wa/3' 

12.  Express  SaiWasW*  in  terms  of  the  st  when  (t)  o>6>c>d>0,  and  (ii)  when 
a  =  &  =  c=d. 

13.  By  solving  the  first  k  of  Newton's  identities  (10)  as  a  system  of  linear  equations, 
find  an  expression  in  the  form  of  a  determinant  (i)  for  st  in  terms  of  ch  .  .  .  ,  a,  and 
(ii)  for  C£  in  terms  of  sh  .  .  .  ,  sj. 

14.  One  set  of  n  numbers  is  a  mere  rearrangement  of  another  set  if  S\,  .  .  .  ,  sn 
have  the  same  values  for  each  set. 


CHAPTER  X 
Elimination,  Resultants  and  Discriminants 

110.  Elimination.     If  the  two  equations 

ax+b  =  0,         cx+d  =  0  (a  9*0,  c=*0) 

are  simultaneous,  i.e.,  if  x  has  the  same  value  in  each,  then 

x= — = — ,  R=ad  —  bc  =  0, 

a         c 

and  conversely.  Hence  a  necessary  and  sufficient  condition  that  the 
equations  have  a  common  root  is  R  =  0.  We  call  R  the  resultant  (or 
eliminant)  of  the  two  equations. 

The  result  of  eliminating  x  between  the  two  equations  might  equally 
well  have  been  written  in  the  form  be—  ad  =  0.  But  the  arbitrary  selec- 
tion of  R  as  the  resultant,  rather  than  the  product  of  R  by  some  constant, 
as  —  1,  is  a  matter  of  more  importance  than  is  apparent  at  first  sight.  For, 
we  seek  a  definite  function  of  the  coefficients  a,  b,  c,  d  of  the  functions 
ax-\-b,  cx-\-d,  and  not  merely  a  property  R  =  0  or  Ry*0  of  the  correspond- 
ing equations.  Accordingly,  we  shall  lay  down  the  definition  in  §  111, 
which,  as  the  reader  may  verify,  leads  to  R  in  our  present  example. 

Methods  of  elimination  which  seem  plausible  often  yield  not  R  itself, 
but  the  product  of  R  by  an  extraneous  function  of  the  coefficients.  This 
point  (illustrated  in  §  114)  indicates  that  the  subject  demands  a  more 
careful  treatment  than  is  often  given. 

111.  Resultant  of  Two  Polynomials  in  x.     Let 

f(x)=a0xm+a1xm-1+  .  .  .  +am  («o^O), 


1  g(x)=b0xn+b1xn-1+  .  .  .  +bn  (60^0) 

be  two  polynomials  of  degrees  m  and  n.  Let  ai,  .  .  .  ,  am  be  the  roots 
of  f(x)=0.  Since  a\  is  a  root  of  g(x)=0  only  when  g(ai)=0,  the  two 
equations  have  a  root  in  common  if  and  only  if  the  product 

g(ai)g(a2)  .  .  .  g{am) 
143 


144  ELIMINATION,  RESULTANTS,  DISCRIMINANTS  [Ch,  X 

is  zero.     This  symmetric  function  of  the  roots  of  f(x)=0  is  of  degree  n 
in  any  one  root  and  hence  is  expressible  as  a  polynomial  of  degree  n  in  the 
elementary  symmetric  functions   (§  104),   which  are  equal  to    —a\/ao, 
G^/ao,  ....     To  be  rid  of  the  denominators  ao,  it  therefore  suffices  to 
multiply  our  polynomial  by  ao11.    We  therefore  define 

(2)  R(f,g)=a0ngMg(a2)  .  .  .  g(am) 

to  be  the  resultant  of  /  and  g.     It  equals  an  integral  rational  function  of 
ao,  .  .  .  ,  am,  bo,  .  .  .  ,  bn  with  integral  coefficients. 

EXERCISES 

1.  If  w  =  l,  n  =  2,     R(f,  g)=b0a12—bia0ai-{-b2ao2. 

2.  If  m  =  2,  n  =  l,     R(f,  g)  =a0(6o«i+bi)(bo«2+W  =a06i2— ai&o&i+o^&o2,  since 

ao(ai-\-a2)  =  —  Oi,  aoaia2=02. 

3.  If  /3i,  .  .  .  ,  ft  are  the  roots  of  g(x)  =  0,  so  that 

g(ai)=&o(a1-J8i)  (aj-ft)  .  .  .  (a<-ft»), 
then 

#(/,  g)=a0nbom(f*i-Pi)  (oa-ft  .  .  .    («i-|8„) 

•(as -ft)  (aj-ft)  .  .  .  (a2-ft,) 


'(«!»— ft) (am—  ft)  .  .  .  (am— ft,). 
Multiplying  together  the  differences  in  each  column,  we  see  that 

#(/,  ?)  =  (-l)CTre&om/(ft)/(ft)  .  .  .f(t3n)  =  (-l)mnR(g,f). 

4.  If  m  =  2,  n  =  l,  R(g,f)  =  bo2/(  —  &1/&0)  =  a0&i2  — a1&o&i+a2feo2,  which  is  equal  to  R(f,  g) 
by  Ex.  2.     This  illustrates  the  final  result  in  Ex.  3. 

5.  If  m=n  =  2,  R(f,  g)=ao2b02ai2a22+ao2bobiaia2(a1-\-a2) 

+a02bob2(ai2+a22)  +ao2&i2«ia2+a02&ife2(«i+a2)  +a02b22 
=  b02O22  —  bobiaia2+bob2(ai2 — 2  a0a2)+6i2a0a2  —  bAooai+ao2^2. 
This  equals  i2(gr,  /),  since  it  is  unaltered  when  the  a's  and  6's  are  interchanged. 

6.  Prove  by  (2)  that  R  is  homogeneous  and  of  total  degree  m  in  60,  .  .  .  ,  bn;  and 
by  Ex.  3,  that  R  is  homogeneous  and  of  total  degree  n  in  a0,  .  .  .  ,  am.  Show  that  R 
has  the  terms  a0nbnm  and  (-l)mn60mamn. 

7.  R(S,  gig*)=R<J,  gi)-R<J,  gi). 

8.  JR(/,  zB)  =  ( -  l)mnR(xn,  /)  =  ( - l)mnamn. 


§112] 


SYLVESTER'S  METHOD  OF  ELIMINATION 


145 


112.  Silvester's  Dialytic  Method  of  Elimination.1     Let  the  equations 

fix)  =aox3+aix2+a2x+a3  =  0,        g(x)  =b0x2-\-b1x+b2  =  0 

have  a  common  root  x.     Multiply  the  first  equation  by  x  and  the  second 
by  x2  and  x  in  turn.     We  now  have  five  equations 

aox4:-\-aix3-\-a2X2+a3X        =  0, 

aox?-\-aix2-\-a,2X-\-a3  =  0, 
b0xi+bix3+b2x2  =0, 

boX3  +  biX2-\-b2X  =  0, 
box2+bix-\-b2  =  0, 

which  are  linear  and  homogeneous  in  x4,  x3,  x2,  x,  1.     Hence  (§  97) 


(3) 


F  = 


ao 

ai 

Cl2 

«3 

0 

a0 

«1 

«2 

bo 

&i 

b2 

0 

0 

bo 

&i 

62 

0 

0 

&o 

61 

0 

«3 

0 
0 

&2 


must  be  zero.  Next,  if  F  =  0,  there  exist  (§97)  values  which,  when 
substituted  for  x4,  re3,  x2,  x  and  1,  satisfy  the  five  equations.  But  why  is 
the  value  for  x4  the  fourth  power  of  the  value  for  x,  that  for  x3  the  cube  of 
the  value  for  x,  etc.?  Since  the  direct  verification  of  these  facts  would 
be  very  laborious,  we  resort  to  a  device  to  show  that,  conversely,  if  F  =  0 
the  two  given  equations  have  a  root  in  common. 

In  (3)  replace  as  by  0,3  —  z  and  consider  the  equation 


(4* 


«0 

ai 

«2 

«3  — 

z   0 

0 

ao 

ai 

d2 

az-z 

60 

bi 

b2 

0 

0 

0 

bo 

bi 

62 

0 

0 

0 

bo 

61 

62 

=  0. 


To  prove  that  it  has  the  roots /(ft)  and/(/32),  where  /3i  and  fa  are  the  roots 
of  g(x)=0,  we  take  z=f(Pi)  and  prove  that  the  determinant  is  then  equal 
to  zero.     For,  if  we  add  to  the  last  column  the  products  of  the  elements 

1  Given  without  proof  by  Sylvester,  Philosophical  Magazine,  1840,  p.  132. 


146 


ELIMINATION,  RESULTANTS,  DISCRIMINANTS 


[Ch.X 


of  the  first  four  columns  by  /3t4,  &3,  j3*2,  ft,  respectively,  we  find  that  ail 
of  the  elements  of  the  new  last  column  are  zero. 
Since  (4)  reduces  to  (3)  for  2  =  0,  it  is  of  the  form 

b03z2+kz+F  =  0, 

in  which  the  value  of  k  is  immaterial.  By  considering  the  product  of 
the  roots  of  this  quadratic  equation,  we  see  that 

^  =  &o3/(/3i)/(/32). 

Hence  the  Sylvester  determinant  F  is   the   resultant  R (g,  /)  and  hence 
is  the  resultant  R(f,  g),  since  mn  is  here  even  (Ex.  3,  §  111). 
In  general,  if  the  equations  are 

f(x)=aoxm+  .  .  .  +am  =  0,         g(x)  =bQxn+  .  .  .  +&»  =  0, 

we  multiply  the  first  equation  by  xn~1,  xn~2,  .  >  .  ,  x,  1,  in  turn,  and  the 
second  by  xm_1,  xm~2,  .  .  .  ,  x,  1,  in  turn.  We  obtain  n-\-m  equations 
which  are  linear  and  homogeneous  in  the  m-\-n  quantities  xm+n~1,  .  .  .  , 
«,  1.     Hence  the  determinant 

do   oi  fl2     .  .  .  am  0 0 

0     ao   ci   a2     .  .  .  am  0 0 

0     0     a0   a\   a2     ...  am  0  ...  0 


(5) 


F  = 


0 


.  0  bo  bi br, 


m  rows 


is  zero.  It  may  be  shown  to  be  equal  to  the  resultant  R(f,  g),  whether 
mn  is  even  or  odd,  by  the  method  employed  in  the  above  case  m  =  3,  n  =  2. 
We  may  also  prove  as  follows  that  if  F  =  0  the  equations  /=0  and 
<7  =  0  have  a  common  root.  Since  F  was  obtained  as  the  determinant 
of  the  coefficients  of 

xn~1f,  .  .  .  ,  xf,f,        xm~lg,  .  .  .  ,  xg,  g, 
F  =  0  implies,  by  §  96,  Lemma  2,  the  existence  of  a  linear  relation 

B0xn-lf+  .  .  .  +Bn_2xf-\-Bn_lf+A0xm-lg+  .  .  .  +Am_2xg+A, 


tf^O, 


§112] 


SYLVESTER'S  METHOD  OF  ELIMINATION 


147 


identically  in  x,  with  constant  coefficients  Bo,  . 
In  other  words,  /3/+ag  =  0,  where 


,  ACT_!  not  all  zero. 


(6)     a=A0xm-1+  .  .  .  +Am-2x+Am_1,    ^=B0xn-1+  .  .  .  +Bn-&+Bn-1. 

Neither  a  nor  /3  is  identically  zero.  For,  if  «=0,  for  example,  then 
/3/==0  and  /3  =  0,  whereas  the  At  and  Bt  are  not  all  zero. 

Consider  the  factored  forms  of  /,  g,  a,  /3.  Suppose  that  /  and  g  have 
no  common  linear  factor.  The  highest  power  of  each  linear  factor  occur- 
ring in  /  divides  ag=  —fif  and  hence  divides  a.  Thus  /  divides  a,  whereas 
/  is  of  higher  degree  than  a.  Hence  our  assumption  that  /=0  and  g  =  0 
have  no  common  root  has  led  to  a  contradiction. 

A  similar  idea  is  involved  in  the  method  of  elimination  due  to  Euler  (1707-1783). 
If  /=0  and  <7  =  0  have  a  common  root  c,  then  f  =  (x— c)a,  —g  =  (x—c)p,  identically  in 
x,  where  a  and  /3  are  polynomials  in  x  of  degrees  m  —  1  and  n  —  1,  respectively.  Give 
them  the  notations  (6).  In  the  identity  j3f+ag=0,  the  coefficient  of  each  power  of  x 
is  zero.     Hence 


aoBo 
aiBo+aoBi 


+b0A0 
+biAo+b0A1 


=0 

=0 


OmBn-2-\-am-iBn-i 
&mtin  —  1 


+bnAm-2+bn-  iAm- 1  =0 
+6„Am_1=0. 


Since  these  m-\-n  linear  homogeneous  equations  in  the  unknowns  B0,  .  .  .  ,  2?n_  i,  A0,  .  .  .  ■ 
Am-i  have  a  set  of  solutions  not  all  zero,  the  determinant  of  the  coefficients  is  zero, 
By  interchanging  the  rows  and  columns,  we  obtain  the  determinant  (5) 

EXERCISES 


1.  For  m=n=2,  show  that  the  resultant  is 

do     CLi        02        0 


R  = 


0     ao     ai     02 

b0   61      62      0 

0     b0      bi      b2 
Interchange  the  second  and  third  rows,  apply  Laplace's  development,  and  prove  that 

R  =  (a0b2) 2  —  (a0bi)  (aib2) , 
where  (0062)  denotes  aob2  —  a2ba,  etc. 


148 


ELIMINATION,  RESULTANTS,  DISCRIMINANTS 


[Ch.X 


a0 

ai 

a3 

a3 

0 

0 

a0 

ax 

a2 

03 

0 

0 

0 

a0 

ai 

a2 

a3 

0 

b0 

&i 

02 

63 

0 

0 

0 

0 

Oo 

ai 

02 

a3 

0 

Oo 

Ol 

a2 

o3 

0 

bo 

&i 

b2 

&3 

0 

0 

0 

b0 

&1 

62 

&3 

0 

0 

bo 

h 

fe2 

&3 

0 

0 

0 

a0 

Ol 

«2 

03 

0 

0 

h 

6i 

&2 

bz 

0 

0 

&0 

Ol 

62 

h 

2.  For  ?n=n  =  S,  write  down  the  resultant  R  and,  by  interchanges  of  rows,  derive 
the  second  determinant  in 


R 


To  the  second  determinant  apply  Laplace's  development,  selecting  minors  from  the 
first  two  rows,  and  to  the  complementary  minors  apply  a  similar  development.  This 
may  be  done  by  inspection  and  the  following  value  of  —  R  will  be  obtained: 

(ao&i)  { (aib2)  (a263)  —  (aib3)  2+ (a2b3)  (a0b3) } 
—  (a062)  { (a062)  (0263)  —  (a0b3)  (aib3) } 
+  (a0b3)  { (ooMfeos)  -(0063)2}  • 

The  third  term  of  the  first  line  and  the  first  term  of  the  last  line  are  alike.  Hence, 
changing  the  signs, 

R  =  (0063) 3 — 2  (o06i)  (aob3)  (a2b3)  —  (o062)  (aO03)  (oi63) 

+  (00&2)  2(a2b3)  +  (oobi)  (oi03) 2  —  (a06i)  (axb2)  (a2b3) . 

Other  methods  of  simplifying  Sylvester's  determinant  (5)  are  given  in  §  113. 

113.  Bezout's  Method  of  Elimination.  When  the  two  equations  are 
of  the  same  degree,  the  method  published  by  Bezout  in  1764  will  be  clear 
from  the  example 


f=a0x3-{-aiX2-{-a2x+a3  =  0,         g  =  b0x3-\-bix2-\-b2x-Jrb3  =  0. 


Then 

(7) 


aog-bof, 
(aQx-\-a{)g  —  (b0x+bi)f, 
(aox2jt-aix+a2)g—(box2+bix+b2)f 
are  equal  respectively  to 

(a0bi)x2  Jr(aob2)x   +((1063)  =0, 

(8)  (aob2)x2+  {(ao&3)  +  (ai&2)}z+(ai&3)=0, 

(a0&3)z2  +(ciib3)x    +(a2b3)=0, 


§113] 


BEZOUT'S  METHOD  OF  ELIMINATION 


149 


where  (ao&i)=ao&i  —  aibo,  etc.  The  determinant  of  the  coefficients  is  the 
negative  of  the  resultant  R(f,  g).  Indeed,  the  negative  of  the  determi- 
nant is  easily  verified  to  have  the  expansion  given  at  the  end  of  Ex.  2 
just  above. 

To  give  a  more  instructive  proof  of  the  last  fact,  note  that,  by  (7),  equations  (8) 
are  linear  combinations  of 

scy=0,         xf=0,        /=0,         x2g  =  0,         xg  =  0,         g  =  0, 

the  latter  being  the  equations  used  in  Sylvester's  method  of  elimination.  The  deter- 
minant of  the  coefficients  in  these  six  equations  is  the  first  determinant  R  in  Ex.  2  just 
above.  The  operations  carried  out  to  obtain  equations  (8)  are  seen  to  correspond 
step  for  step  to  the  following  operations  on  determinants.  To  the  products  of  the  ele- 
ments of  the  fourth  row  by  a0  add  the  products  of  the  elements  of  the  1st,  2nd,  3rd, 
5th,  6th  rows  by  —  b0,  —  &i,  —  b2,  a.\,  02  respectively  [corresponding  to  the  formation 
of  the  third  function  (7)].  To  the  products  of  the  elements  of  the  fifth  row  by  a0  add 
the  products  of  the  elements  of  the  2nd,  3rd,  6th  rows  by  —  bo,  —  &i,  «i  respectively  [cor- 
responding to  the  second  function  (7)].  Finally,  to  the  products  of  the  elements  of  the 
sixth  row  by  ao  add  the  products  of  the  elements  of  the  third  row  by  —  bQ  [corresponding 
to  a0g  —  b0f\.     Hence 

a0     ai     02  a3  0  0 


a03R  = 


so  that  R  is  equal  to  the  3-rowed  minor  enclosed  by  the  dots.  The  method  of  Bezout 
therefore  suggests  a  definite  process  for  the  reduction  of  Sylvester's  determinant  of 
order  2n  (when  m=n)  to  one  of  order  n. 

Next,  for  equations  of  different  degrees,  consider  the  example 


0 

a0 

ai 

«2 

a3 

0 

0 

0 

a0 

eti 

ch 

03 

0 

0 

0 

(a063) 

(0163) 

(a2b3) 

0 

0 

0 

(00&2) 

(ao&s)  +  (ai&2) 

(01&3) 

0 

0 

0 

(aobi) 

(ao&2) 

(a0b3) 

Then 


f=aoXiJra\X3-\-aix'2JraiX-irai)         g  =  &0£24-&i£+&2- 
aox2g—b0f,         (a0x+ai)x2g  —  (boX+b^f 


are  equal  respectively  to 

(a0bi)x3  +  (aob2)x2 — a3boX— aj}0, 

(a0b2)x3  +  { (aib2)  —a3b0 }  x2  —  { a3&i  +a4&o }  * — a4bi . 

The  determinant  of  the  coefficients  of  x3,  x2,  x,  1  in  these  two  functions  and  xg,  g,  after 
the  first  and  second  rows  are  interchanged,  is  the  determinant  of  order  4  enclosed  by 
flots  in  the  second  determinant  below.     Hence  it  is  the  resultant  R(f,  g) . 


150 


ELIMINATION,  RESULTANTS,  DISCRIMINANTS 


[Ch.  X 


Clo 

«1 

Ch 

a3 

04 

0 

0 

ao 

ai 

CH 

a3 

a4 

&o 

by 

b2 

0 

0 

0 

0 

bo 

hi 

b2 

0 

0 

0 

0 

bo 

h 

b2 

0 

0 

0 

0 

bo 

bi 

&2 

As  in  the  former  example,  we  shall  indicate  the  corresponding  operations  on  Syl- 
vester's determinant 


R 


Multiply  the  elements  of  the  third  and  fourth  rows  by  ao.  In  the  resulting  determinant 
a02R,  add  to  the  elements  of  the  third  row  the  products  of  the  elements  of  the  first, 
second  and  fourth  rows  by  —  bo,  —  bi,  ai/a0  respectively.  Add  to  the  elements  of  the 
fourth  row  the  products  of  those  of  the  second  by  —  b0.     We  get 


a02R  = 


a0 

ai 

Ch 

a3 

04 

0 

0 

ao 

ai 

a2 

a3 

a^ 

0 

0 

(a0&2) 

(aib2)  — a3bo 

— a36i—  aib0 

—aj)i 

0 

0 

(a0bi) 

(a0&2) 

— a3b0 

— a46o 

0 

0 

6o 

&i 

h 

0 

0 

0 

0 

bo 

6i 

h 

Hence  R  is  equal  to  the  minor  enclosed  by  dots. 


EXERCISES 


1.  For  to  =  3,  n  =  2,  apply  to  Sylvester's  determinant  R  exactly  the  same  operations 
as  used  in  the  last  case  in  §  113  and  obtain 


R- 


{ciob2)      (aib2)  —  a36o      —  a3bi 

(a0&i)  (ciob2)  —  a3b0 

b0  bi  b2 


2.  For  TO  =  n=4,  reduce  Sylvester's  R  (as  in  the  first  case  in  §  113)  to 

(a0bi)  (a0b2)  (aoh)  (aob4) 

(a0b2)  (a0b3)  +  (aib2)  (a0b4)  +  (aib3)  (aib4) 

(a0b3)  (a0b4)-r-(aib3)  (aib4)  +  (a2b3)  {a2b^) 

(aoh)  (aib4)  (aiib4)  (a3bi) 


§114] 


GENERAL  THEOREM  ON  ELIMINATION 


151 


114.  General  Theorem  on  Elimination.  If  any  method  of  eliminating 
x  between  two  equations  in  x  leads  to  a  relation  F  =  0,  where  F  is  a  polynomial 
in  the  coefficients,  then  F  has  as  a  factor  the  true  resultant  of  the  equations. 

Some  of  the  preceding  proofs  become  simpler  if  this  theorem  is  applied. 
For  example,  determinant  (3)  is  divisible  by  the  resultant  R.  Since  the 
diagonal  term  of  (3)  is  a  term  ao2b23  of  R  (Ex.  6,  §  111),  i^  is  identical 
with  R. 

The  preceding  general  theorem  is  proved  in  the  author's  Elementary 
Theory  of  Equations,  pp.  152-4.  We  shall  here  merely  verify  the  theorem 
in  an  instructive  special  case.     Let 

f=aox3+aix2+a2X+a3  =  0,         g=box3-\-biX2-\-b2x-\-b3  =  0 

have  a  common  root  x^O.     Then 

—  bof-\-aog  =  (a06i)x2+  (a0b2)x+  (0063), 

(63/-  a3g)/x  =  (aob3)x2+  (aib3)x+  (0263) . 

By  Ex.  1  of  §  112,  the  resultant  of  these  two  quadratic  functions  is 


F  = 


(a063)     (ao&i)    2_      (aob3)   (ao&i)  (ai&s)   (a0b2) 

(a263)     (a0fr3)  (0163)   (ao&2)  (02^3)   (0063) 

This  is,  however,  not  the  resultant  R  of  the  cubic  functions  /,  g.  To  show 
that  (aob3)  is  an  extraneous  factor,  note  that  the  terms  of  F  not  having 
this  factor  explicitly  are 

(ao&i)  (a2&3)  { (ao&i)  (a2&3)  —  (aob2)  (0163) } . 

The  quantity  in  brackets  is  equal  to  —  (ao&3)(ai&2),  since,  as  in  Ex.  2 
of  §  101, 


0=4 


ao 

a\ 

a2 

as 

bo 

61 

62 

b3 

ao 

a\ 

«2 

as 

bo 

bi 

b2 

b3 

t/zo&i)  (a2b3)  -  (0062)  (ai63)  +  (0063)  (ai&2) . 


We  now  see  that  F=(a0b3)R,  where  R  is  given  in  Ex.  2  of  §  112.  This 
method  of  elimination  therefore  introduces  an  extraneous  factor  (0063). 
The  student  should  employ  only  methods  of  elimination  (such  as  those 
due  to  Sylvester,  Euler,  and  Bezout)  which  have  been  proved  to  lead 
to  the  true  resultant. 


152  ELIMINATION,  RESULTANTS  AND  DISCRIMINANTS        [Ch.  X 

EXERCISES 

Find  the  result  of  eliminating  x  and  hence  find  all  sets  of  common  solutions  of 
1.  x2-y2  =  9,  xy  =  5y.      2.  x2+y2  =  25,  x2+3(c-l)x+c(y*-25)  =0. 

3.  When  x2+ax+b=0  has  a  double  root,  what  3  rowed  determinant  is  zero? 

4.  Find  the  roots  of  x6+3x4+32z3+67£2+32x+65  =  0  by  §  79. 

115.  Discriminants.     Let  a\,  .  .    . ,  am  be  the  roots  of 

(9)  f(x)  =a0xm+alxm- l+  .  .  .  +a„,  =  0  (a0^0), 
so  that 

(10)  f(x)  =a0(x-a1)(x-a2)  .  .  .  (x-am). 
As  in  §  44,  we  define  the  discriminant  of  (9)  to  be 

D  =  ao2m~2(ai-a2)2(a]_-a3)2  .  .  .  (ai-am)2(a2-a3)2  .  .  .  (am_i— am)2. 

Evidently  D  is  unaltered  by  the  interchange  of  any  two  roots.  Since  the 
degree  in  any  root  is  2{m—l);  the  symmetric  function  D  is  equal  to  a 
polynomial  in  ao,  .  .  .  ,  am.  Indeed,  ao2m~2  is  the  lowest  power  of  ao 
sufficient  to  cancel  the  denominators  introduced  by  replacing  2ai  by 
—  ai/ao,  .  .  .  ,  a\a2  ...  am  by  ±am/ao.  By  differentiating  (10),  we  see 
that 

f(ai)=ao(ai—  0L2) (0:1—0:3)  •  •  •  (ai—am), 
f'{a2)=ao{a2—ai){(x2—az)  .  .  .  (a2—am), 
f'(a3)=ao(a3—ai) (0:3—0:2) (0:3—  0:4)  ■  •  ■  (0:3—  am), 

etc.     Hence 

oom-7'(«i)  •  •  ./,W=ao2m-1("l)1+2+  •  •  '  +ro-1(«i-«2)2  •  •  •  (owi-og2 

m(m  —  1) 

=  (-1)     2      aoD. 
By  (2),  the  left  member  is  the  resultant  of  f(x),  f'(x).     Henc'e 


m(m  —  1) 


1 


(id  d=(-d  2  -R(f,n 

ao 


§115] 


DISCRIMINANTS 


153 


EXERCISES 

1.  Show  that  the  discriminant  of  f=y3+py-\-q  =  0  is  —  4p3  — 27  g2  by  evaluating 
the  determinant  of  order  five  for  R(J,  /'). 

2.  Prove  that  the  discriminant  of  the  product  of  two  functions  is  equal  to  the  prod- 
uct of  their  discriminants  multiplied  by  the  square  of  their  resultant.  Hint:  use 
the  expressions  in  terms  of  the  differences  of  the  roots. 

3.  For  00  =  1,  show  that  the  discriminant  is  equal  to 


So 


Sm-l 


sm 


1        am     am'    .  .  .  am'  Sw_i       Sm        Sm+i     .  .  .   S2m-2 

where  Si=ail+  .  .  .  +am\     See  Ex.  4,  §  88;  Ex.  2,  §  102. 

4.  Hence  verify  that  the  discriminant  of  x3+px+g  =  0  is  equal  to 
3  0       -2p 
0       -2p     -3q       =-4p3-27g2. 

-2p     -Sq        2p2 

5.  By  means  of  Ex.  1,  §  113,  show  that  the  discriminant  of  a0x3+aiX2+a2X-\-a3  =  0  is 

2ao«2    ai02+3ffloa3    2ai03 

ai  2a2  3o3 

3ao  2ai  a^ 

=  18a0aia2O3  — 4a0O23— 4ai3a3+ai2a22— 27a02a32. 


MISCELLANEOUS   EXERCISES 

1.  Find  the  equation  whose  roots  are  the  abscissas  of  the  points  of  intersection 
of  two  general  conies. 

2.  Find  a  necessary  and  sufficient  condition  that 

f(x)  =x4-\-px3-\-qx2-\-rx-\-s  =  0 

shall  have  one  root  the  negative  of  another  root.  When  this  condition  is  satisfied, 
what  are  the  quadratic  factors  of /(x)?  Apply  to  Ex.  4,  §  74.  Hint:  add  and  subtract 
f{x)  and/(—  x). 

3.  Solve  /(x)  =x4— 6x3+13z2  — 14x+6  =  0,  given  that  two  roots  a  and  0  are  such 
that  2a+/3  =  5.     Hint:  /(x)  and/(5— 2x)  have  a  common  factor. 

4.  Solve  x3+px+q  =  0  by  eliminating  x  between  it  and  x2-\-vx-\-w  =  y  by  the  greatest 
common  divisor  process,  and  choosing  v  and  w  so  that  in  the  resulting  cubic  equation 
for  y  the  coefficients  of  y  and  y2  are  zero.     The  next  to  the  last  step  of  the  elimination 


154  ELIMINATION,  RESULTANTS  AND  DISCRIMINANTS       [Ch.  X 

gives  x  as  a  rational  function  of  y.     (Tschirnhausen,  Acta  Erudit.,  Lipsiae,  II,  1683, 
p.  204.) 

5.  Find  the  preceding  y-cubic  as  follows.  Multiply  x2-\-vx-\-w  =  y  by  x  and  replace 
x3  by  —  px  —  q;  then  multiply  the  resulting  quadratic  equation  in  x  by  x  and  replace 
x3  by  its  value.     The  determinant  of  the  coefficients  of  x2,  x,  1  must  vanish. 

6.  Eliminate  y  between  y3=v,    x  —  ry-^-sy2,  and  get 

x3— Zrsvx  —  {r3v-\-s3v2)  =0. 

Take  s  =  l  and  chose  r  and  v  so  that  this  equation  shall  be  identical  with  x3+px+q  =  0, 
and  hence  solve  the  latter.     (Euler,  1764.) 

7.  Eliminate  y  between  y3  =  v,     x=f+ey-\-y2  and  get 


1  e       f—x 

e       f—x        v 
f—x        v  ev 


=  0. 


This  cubic  equation  in  x  may  be  identified  with  the  general  cubic  equation  by  choice 
of  e,  f,  v,     Hence  solve  the  latter. 

8.  Determine  r,  s  and  v  so  that  the  resultant  of 

x-\-r 

y3=v,       y=—- 
y+s 

shall  be  identical  with  x3-\-px+q  =  0.     (Bezout,  1762.) 

9.  Show  that  the  reduction  of  a  cubic  equation  in  x  to  the  form  y3  =  v  by  the  sub- 
stitution 

r+sy 

x= 

1+1/ 

is  not  essentially  different  from  the  method  of  Ex.  7.  [Multiply  the  numerator  and 
denominator  of  x  by  1—  y+y2.] 

10.  Prove  that  the  equation  whose  roots  are  the  n(n  —  1)  differences  xj—xt  of  the 
roots  of  f(x)  =0  may  be  obtained  by  ehminating  x  between  the  latter  and  f{x-\-y)  =0 
and  deleting  from  the  eliminant  the  factor  yn  (arising  from  y  =  xj—Xj  =  0).  The 
equation  free  of  this  factor  may  be  obtained  by  eliminating  x  between  f(x)  =  0  and 

n-l 

{f(x+y)-f(x)} /y=f(x)+f"{x)  ^  +.  .  .  +fn)(x)^ -=0. 

This  eliminant  involves  only  even  powers  of  y,  so  that  if  we  set  y2  =  z  we  obtain  an 
equation  in  z  having  as  its  roots  the  squares  of  the  differences  of  the  roots  of  fix)  =  0. 
(Lagrange  Resolution  des  equations,  1798,  §  8.) 

11.  Compute  by  Ex.  10  the  z-equation  when/(x)  =x3+px+q. 


APPENDIX 


THE  FUNDAMENTAL  THEOREM   OF  ALGEBRA 

Theorem.     An  equation  of  degree  n  with  any  complex  coefficients 

f{z)^zn+a1zn~1+  .  .  .  -ra„  =  0 

has  a  complex  (real  or  imaginary)  root. 

Write  z  =  x-\-iy  where  x  and  y  are  real,  and  similarly  a\  =  ci-\-id\,  etc. 
By  means  of  the  binomial  theorem,  we  may  express  any  power  of  z  in  the 
form  X+iY.     Hence 

(1)  f(z)  =  <j>(x,y)+ixP(x,y), 

where  4>  and  \p  are  polynomials  with  real  coefficients. 

The  first  proof  of  the  fundamental  theorem  was  given  by  Gauss  in 
1799  and  simplified  by  him  in  1849.  This  simplified  proof  consists  in 
showing  that  the  two  curves  represented  by  cj>(x,  y)=0  and  \l/(x,  t/)=0 
have  at  least  one  point  {x\,  y\)  in  common,  so  that  zx=xi-\-iyi  is  a  root 
of  f(z)=0.  This  proof  is  given  in  Chapter  V  of  the  author's  Elementary 
Theory  of  Equations. 

We  here  give  a  shorter  proof,  the  initial  idea  of  which  was  suggested, 
but  not  fully  developed,  by  Cauchy.1 

Lemma  1.  aih-\-a2h2-\-  .  .  .  -\-anhn  is  less  in  absolute  value  than  any 
assigned  positive  number  p  for  all  complex  values  of  h  sufficiently  small  in 
absolute  value. 

The  proof  differs  from  that  of  the  auxiliary  theorem  in  §  62  only  in 
reading  "  in  absolute  value  "  for  "  numerically." 

We  shall  employ  the  notation  \z\  for  the  absolute  value  +  Vx2+y2  of 
z  =  x-\-iy. 

1  For  a  history  of  the  fundamental  theorem,  see  Encyclopedic  des  sciences  mathe- 
maliques,  tome  I,  vol.  II,  pp.  189-205. 

155 


156  APPENDIX 

Lemma  2.  Given  any  positive  number  P,  we  can  find  a  positive  number 
R  such  that  \  f(z)  \>Pif\z\^R. 

The  proof  is  analogous  to  that  in  §  64.     We  have 

f(z)=zn(l+D),        Z)-=ai0)+  .  .  .  +a»0)" 

Since  (Ex.  5,  §  8)  the  absolute  value  of  a  sum  of  two  complex  numbers 
is  equal  to  or  greater  than  the  difference  of  their  absolute  values,  we  have 

\f(z)\^\z\n[l-\D\}. 

Let  p  be  any  assigned  positive  number  <1.  Applying  Lemma  1  with 
h  replaced  by  1/z,  we  see  that  |  D  \<p  if  |  1/z  |  is  sufficiently  small,  i.e., 
if  p=|  z  |  is  sufficiently  large.     Then 

\S{z)\>p\l-V)>P 

if  pn^P/(l-p),  which  is  true  if 


*fe 


iR. 


V 
This  proves  Lemma  2. 

Lemma  3.     Given  a  complex  number  a  such  that  f(a)  3^0,  we  can  find 
a  complex  number  z  for  which  \  f(z)  \  <  |  /(a)  | . 

Write  z  =  a+h.    By  Taylor's  theorem  (8)  of  §  56, 

hr  hn 

,^-H...+/w(a).% 

r!  n! 


f(a+h)=f(a)+f'W+  -  ■  ■  +/(r)(«)-^+  •  •  •  +fw(fl)- 


Not  all  of  the  values  f'(a),  /"(a),  ...  are  zero  since  fin)(a)=nl  Let 
/(r)(a)  be  the  first  one  of  these  values  which  is  not  zero.     Then 

/(a+ft)     ,  ,/w(a)  V  ,  ,  /(w)(q)  £ 

/(a)  "^/(a)  "rl-"  "*"/(<»)     nV 

Writing  the  second  member  in  the  simpler  notation 

g(h)  =  l+bhr+chr+1+  .  .  .  +lhn,        b^O, 

we  shall  prove  that  a  complex  value  of  h  may  be  found  such  that  \  g(h)  |  <  1. 
Then  the  absolute  value  of  f(z)/f  (a)  will  be  <1  and  Lemma  3  proved. 
To  find  such  a  value  of  h,  write  h  and  b  in  their  trigonometric  forms  (§  4) 

h  =  p(cos  d+isin  6),         b=\  b  |  (cos /3+t  sin /3). 


FUNDAMENTAL  THEOREM  OF  ALGEBRA  157 

Then  by  §  5,  §  7, 

bhr=\b\pT{  cos  (B+rd)+ism  (/3+r0)}. 

Since  h  is  at  our  choice,  p  and  angle  6  are  at  our  choice.  We  choose  8 
so  that  /3+r0=18O°.  Then  the  quantity  in  brackets  reduces  to  —  1, 
whence 

0(70  =  (1-|  b  \pT)+hr(ch+  .  .  .  +lhn~T). 

By  Lemma  1,  we  may  choose  p  so  small  that 

\ch+  .  .  .  +lhn~r  \<\b\. 

By  taking  p  still  smaller  if  necessary,  we  may  assume  at  the  same  time 
that  |  6  |pr<l.     Then 

\g(h)\<(l-\b\P')  +  Pr\b\,         \g(h)\  <1. 

Minimum  Value  of  a  Continuous  Function.  Let  F(x)  be  any  poly- 
nomial with  real  coefficients:  Among  the  real  values  of  x  for  which 
2£x£3,  there  is  at  least  one  value  x\  for  which  F{x)  takes  its  minimum 
value  F(x\),  i.e.,  for  which  F{x\)  ±F{x)  for  all  real  values  of  x  such  that 
2.£z£3.  This  becomes  intuitive  geometrically.  The  portion  of  the 
graph  of  y  =  F(x)  which  extends  from  its  point  with  the  abscissa  2  to  its 
point  with  the  abscissa  3  either  has  a  lowest  point  or  else  has  several 
equally  low  points,  each  lower  than  all  the  remaining  points.  The  arith- 
metic proof  depends  upon  the  fact  that  F{x)  is  continuous  for  each  x 
between  2  and  3  inclusive  (§  62).  The  proof  is  rather  delicate  and  is 
omitted  since  the  theorem  for  functions  of  one  variable  x  is  mentioned 
here  only  by  way  of  introduction  to  our  case  of  functions  of  two  variables. 

We  are  interested  in  the  analogous  question  for 

G(x,y)  =  <t>2(x,y)  +  e(x,y), 

which,  by  (1),  is  the  square  of  |  f(z)  \.  As  in  the  elements  of  solid  analytic 
geometry,  consider  the  surface  represented  by  Z  =  G(x,  y)  and  the  right 
circular  cylinder  x2-\-y2  =  R2.  Of  the  points  on  the  first  surface  and  on 
or  within  their  curve  of  intersection  there  is  a  lowest  point  or  there  are 
several  equally  low  lowest  points,  possibly  an  infinite  number  of  them. 
Expressed  arithmetically,  among  all  the  pairs  of  real  numbers  x,  y  for 


158  APPENDIX 

which  x2-\-y2  ^.R2,  there  is1  at  least  one  pair  x\,  y\  for  which  the 
polynomial  G(x,  y)  takes  a  minimum  value  G(x±,  y{),  i.e.,  for  which 
G(xi,  y\)  =G{x,  y)  for  all  pairs  of  real  numbers  x,  y  for  which  x2-\-y2  SR2. 

Proof  of  the  Fundamental  Theorem.  Let  z'  denote  any  complex 
number  for  which  f(z')  ^0.  Let  P  denote  any  positive  number  exceeding 
|  f(z')  \.  Determine  R  as  in  Lemma  2.  In  it  the  condition  |  z  \  hR  may 
be  interpreted  geometrically  to  imply  that  the  point  (x,  y)  representing 
z  =  x-\-iy  is  outside  or  on  the  circle  C  having  the  equation  x2-\-y2  =  R2. 
Lemma  2  thus  states  that,  if  z  is  represented  by  any  point  outside  or  on 
the  circle  C,  then  \f(z)  \>P.  In  other  words,  if  |/(z)  \=P,  the  point 
representing  z  is  inside  circle  C.  In  particular,  the  point  representing  z' 
is  inside  circle  C. 

In  view  of  the  preceding  section  on  minimum  value,  we  have 

G(xh  Vl)  <  G(x,  y) 

for  all  pairs  of  real  numbers  x,  y  for  which  x2-\-y2  ^.R2,  where  x\,  y\  is  one 
such  pair.     Write  z\  for  x\-\-iy\.     Since  |  f(z)  \2  =  G(x,  y),  we  have 

for  all  z's  represented  by  points  on  or  within  circle  C.  Since  z'  is  repre- 
sented by  such  a  point, 

(2)  \f(.zi)\^\m\<p- 

This  number  z\  is  a  root  of  f{z)  =0.  For,  if  f(z{)  ^0,  Lemma  3  shows 
that  there  would  exist  a  complex  number  z  for  which 

(3)  l/(z)l<l/0i)l- 

Then  |  f{z)  \<P  by  (2),  so  that  the  point  representing  z  is  inside  circle  C, 

as  shown  above.     By  the  statement  preceding  (2), 

IM)  l±l/(*)l- 

But  this  contradicts  (3).     Hence  the  fundamental  theorem  is  proved. 

1  Harkness  and  Morley,  Introduction  to  the  Theory  of  Analytic  Functions,  p.  79, 
prove  that  a  real  function  of  two  variables  which  is  continuous  thoroughout  a  closed 
region  has  a  minimum  value  at  some  point  of  the  region, 


ANSWERS 


Pages  2,  3 
i.  Si.  2.  2.  3.   -20+20*.      4.   -f.        5.  (8+2V3J. 

6.  W+V-6)+U2VE-B)i.         7.  =i+f3i.       8.  ^+0-^.    .o.  Yes. 
13.  3,  4 and -3,  -4.       14.   ±(5+6t).      15.   ±(3-20.      16.   ±[c+d  +  (c-d)i]. 

Pages  6,  7 

2.  -3,  -3a>,  -3w2;  i,  coi;  w2i;  fl  =  cos40°-H  sin  40°,  wR,  w2R. 

3.  ±(l+t)/V2;    ±(l-i)/V2;    ±  co2. 

Page  9 

4.  -1,'cos  A+i  sin  A  (A  =36°,  108°,  252°,  324°).  6.  R*,  R«,  R*. 

Page  10 

5.  p(p-l).  6.  (p-l)(g-l)(r-l)  if  n=pqr. 

Page  13 
1.  51.  2.  13. 

Page  15 

1.  Rem.  11,  quot.  x2+5x+8.  2.    -61,  2x"-4x3+7a:2-14a:+30. 

3.  -0.050671,  x2+6.09x  +  10.5481.        4.  z2-x-6,  z+2;  4,3,  -2. 

5.  x2-z-6=0,  3,  -2.  6.  2±V5.  7.  2x2-x+2.  8.  jc«+1. 

Page  17 

1.  x3-3x2+2a;=0.  2.  x4-5x2+4  =0.  3.  x4-18x2+81  =0. 

4.  x4-5x3+922-7:c+2=0.         5.  62=4ac.  7.  By  theorem  in  §18. 

159 


160  ANSWERS 

Page  19 
1.  x3-6x2+llx-6=0.  2.  x4-8x2+16=0. 

3.  1,  2.  5.  4,  f,  -f .  6.  1,  3,  5.  7.  1,  1,  1,  3. 

8.  2,  -6,  18.  9-   -3,  1,  5.  10.  5,  2,  -1,  -4. 

11.  2/2-(p2-2g)y+g2=0.  12.  y2-(p3-3pq)y+q3=0. 

13.  ft)  2/2-Z/(P3-3pg)/g+?=0.  14.  P3r=q3. 
(u)  y2-q(p2-2q)y+q*=0.  15.  2,4,  -6. 
(m)  2/2-(p+p/?)2/+2+g+l/g=0. 

Page  20 

1.  5,  - 1  ±  V^3.  2.  1  ±i,  1  ±V2. 

3.  x3-7x2+19a;-13=0.  4-  4,  1  -  V-5,  x3-6x2+14x-24=0. 

6.  ±1,  2±V3.  7.   V3,  2±i.  9-  x3-fx2-fx+-|=0. 
10.  2  +  V3,  x2+2x+2=0.       11,12.  Not  necessarily.      13.  No. 

Page  23 
1.  19J,  3.  2.  6.  3.  2.  4.  3.  5-  0,  -7,  -£. 

Page  25 

1.   -1,  -1,  -6.        2.   -2,  3,  4.        3.  1,  3,  6.        4.    -2,  -4.        5-  None. 

Page  27 
1.  2,  -1,  -4,  5.         2.  9.         3-  8,  9.         4.    -12,  -35.         5.  2,  2,  -3. 

Page  28 

I.     1,  6,   »,    3.        2.     1,    2,    3-       3-  6-        4"     2)  4>  4-        £>•     4,  4)    6-        "•  2>    3}    4- 

7.  a,  8.  f.  10.  x2-12x-12=0.  n.  x3-3x2-12x+54=0. 

Page  30 
i.l,4.     2.   -1,-4.    3.0.7,-5.7.        4.    -0.7,5.7.     5-2,2.     6.  Imaginary. 

Page  40 

5.  x*+x*-4x3-3x2+3x+l=0.  6.    -|(l±V-3),  £(7±V45). 

10.  See  (11),  §32. 

11.  Edges  roots  of  x3-7x2  +  12x-t>   =0,  all  real  (§45)  and  irrational. 

14.  A  =  area,    c=hypotenuse,    squares   of  legs  §(c2±  Vc4-16A2). 

15.  A  area,  a,  &  given   sides,  square   third   side  is  a2+62±2\/a2&2  —  4A2. 

16.  yi-2y3  +  (2-g2)y2-2y+l=0)  pos.  roots  0.09125,  10.95862. 


<z 


ANSWERS  16] 

Page  44 

3.  0=2,  R+R*+R12+R*,  etc.,  z3+z2-4z+l  =0.  _ 

4.  0=2,  i2+fl*,  Rt+R^+R5.  5-  |(l±V-3),  |(-5±V21). 
6.   -1,  2±V3,  ^±§V-3. 7-  1,  1,  1,  -1,  |(1±V-15). 

8.  -1,  -2,  -i  |(-5±V_11). 

Page  46  ^ 

1.    -5,  |(5±Vl3).      2.    _6,  ±V^3.     3.    -2,l±i.      4.  |i(- 2±VZ3). 

Page  48 

1.  A  =  -400,  one.       2.  A  =4 -27 -121,  three.      3.  A  =0,  two.      4.  A  =0,  two. 

Page  49 

1.    -4,  2±  V3.  2.  See  Ex.  1,  §47. 

3.  1.3569  4.   -1.201639  5.       1.24698       6.       1.1642 
1.6920                     1.330058  -1.80194  -1.7729 

-3.0489.  -3.128419.  -0.44504.  -3.3914. 

Page  51 

1.  1,-1,_4±V6.  2.    -1,  -2,2,3.  3.  l±i,  -1±^2. 

4.  1±V2,  -l±V-2.  5-  4,  -2,  -l±i. 

Page  54  (bottom) 

1.  (-3,9).  2.  A=  -250000,  x  =3,  -2,  ±i. 

3.  (3,9),  (-2,4).  4-  ^  =3.  5-  6.856,7. 

Pages  59,  60 

2.  2.1.  3.  (-0.845,  4.921),  (-3.155,  11.079);  between  -4  and  -5. 

4.  1.1,  -1.3  5.  Between  0  and  1,  0  and  -1,  2.5  and  3,  -2.5  and  -3. 

9.  120(x3+x),  120x2-42. 

Page  62 

1.  3.      2.  2,  -2.     3.    -1.     4.  Double  roots,  1,3.     5.  None.     6.  3,3,  -3,6. 

Pages  64,  65 

3.  Use  Ex.  3,  p.  62,  abscissas  -1,  3.      4.  Use  Ex.  2,  p.  62. 
6.  i/=-15x-7,  X3-15X+23=0. 


162  ANSWERS 

Page  66 

i.  One  real.  2.  (iV^,  7=Fi^V|-),  three  real. 

3.  (±Vf,  -l=F£Vf),  three.  4.  (-2±V5,  23=F10V5),  one. 

Pages  74,  75 

13.  y5+2yi+5y3+3y2-2y-9=0.  14.  y3  +  15?/2+52?/-36=0. 

Page  78 

1.  One,  between  —2  and  —3.  2.  One,  between  1  and  2. 

Pages  79,  80 

1.  (-4,  -3),  (-2,  -1),  (1,  2).  2.  (-2,-1),  (0,  1). 
3.  (-2,  -1.5),  (-1.5,  -1),  (3,  4).       4.  (-2,  -1),  (0,  1). 
5.  (-7,  -6),  (1,2).  6.  (0,1),  (3,4). 

Page  83 

2.  1,  1,  1,  2.  3.  1,  1,  —2,  —2.  4.  1,  1,  two  imaginary. 

Page  85 

1.  (-2,  -1),  (0,  1),  (1,  2).  2.   (-4,  -3),  (-2,  -1),  (1,  2). 

Pages  89,  90 

1.  Single,  -2.46955.  2.    -1.20164,  1.33006,  -3.12842. 

3.  1.24698,  -1.80194,  -0.44504.  4.   ±2.1213203,  2.1231056,  -6.1231056. 
5.  3.45592,  21.43067.  6.  2.15443. 

7.  -1.7728656,  1.1642479,  -3.3913823. 

8.  3.0489173,  -1.3568958,  -1.6920215. 

9.  2.24004099.  10.  1.997997997. 
11.  1.094551482.                                       12.  2.059,  -1.228. 

13.  1.2261.  14.  0.6527  =  reciprocal  of  2  cos  40°. 

15.  0.9397.  16.  1.3500.  17.  2.7138,  3.3840. 

18.  5.46%.  19.  5.57%.  20.  9.70%. 

Page  94 
1.  2.  2.  3. 


ANSWERS  163 

Page  96 
1.  2.24004099.  2.  2.3593041.  3.  1.997998. 

Page  98 
1.  132°  20.7'.  2.  157°  12'.  3-  4.8425364.  4.  3.1668771. 

5,  7.  15°  16|'}  85°  56|',  212°  49',  225°  57'. 

6.  72°  17'.  8.  5°  56*',  25°  18'.        9.  2.5541949.  10.  1.85718. 

Page  99 
1.    -1.04727 ±1.13594 1.  2.    -f±-|V3i. 

3.   -l±i.  4.  l±i,  l±2i.  5.  2±£,  ±2i. 

Page  100 

1.  217°  12'  27.4"  =3.790988  radians.      2.  42°  20'  47|"  doubled. 

3.  133°  33.8'.  4.  108°  36'  14".  5.  21.468212. 

6.  Angle  at  center  47°  39'  13".  7.  49°  17'  36.5". 

8.  1.4303  x,  2.4590  w,  3.4709  tt;    257°  27'  12.225"  more  exact  than  first. 

9.  x/w  =0.6625,  1.891,  2.930,  3.948,  4.959. 

10.  (i)  0.327739,  0.339224,  1124.333037. 

(u)  0.250279,  0.894609,  1.127839.     [Set  x  =  1  +y,  y  =  1/z  and  solve  by  trigo- 
nometry.] 

11.  3.597285.  12.  10,  1.371288. 
13.  0.  326878,  12.267305.                         14.  324°  16'  29.55". 

15.  10  yr.  4  mo.  0  days.  16.  6.074674.  17.  6.13%. 

Page   102 
i.x=5,y=6.  2.x=2,y  =  l.  ^.x=a,y=0. 

Page   106 

I.   —a2biCidi+a2biCidi+a2b3Cidi— 02^0^1— a2biCid3+a2b4C3d1.      2.   +,  +. 

Page   112 
3.   -3.  4-    -8. 

Page   115 

1.  x  =  -8,  y  =  -7,  z  =26.  2.  x=3,  y  =  -5,  z  =2. 

3.  x=6,  y=3,  z  =  12.  4.  x  =5,  y  =4.  2=3. 

5.  x=  -5,  y=3,  2=2,  w  =  l.  6.  x  =  l,  ?/=z=0,  w=  —  1. 


164  ANSWERS 

Page   119 

1.  Consistent:  y  =  —8/7—  2x,  2=5/7  (common  line). 

2.  Inconsistent,  case  (/3).  3.  Inconsistent  (two  parallel  planes). 

4.  Consistent  (single  plane). 

a-1  -3 

5.  00  z=  —  x—  y  —  2;   (ri)  inconsistent;    (iti)  a;  = — — ,  y=z=— — — . 

,    ,_         (k-b)(c-k)      ....        k-c  a-k-ti  7  u  *   • 

6.  (z)  z  =  - — -;     (it)  w  = x,  z  = if  k=a  or  k  =  c,   but  mcon- 

(a-6)(c-a)  a-c  a-c 

sistent  if  A;  is  different  from  a  and  c;   (ro)  2  =  l-z-?/if/<;=a,  inconsistent  if  k  9^  a. 


Page  120 

1.  r  =  2,x  :y  :z=  -4  : 1  : 1.  2.  r=2,  x  :y  :z=  -10:  8  :  7. 

3.  r  =  1,  two  unknowns  arbitrary.  4.  r  =3,  a;  :  y  :  2  :  w  =6  :  3  :  12  :1. 

5.  r  =  2,    z  =  -^x  — ^y,    w  =  - ±£x  -±gy. 

Page  121 


1.  Ranks  of  A  and  5  are  2;    y=  -8/7  -2x,  2=5/7. 

-5 
61' y    QV" 


—5         3         45 

2.  Consistent  only  when  a  =  —225/61  and  then  x  =  — -,  y  =  — ,  2  =- 


3.  Rank  of  A  is  2,  rank  of  J5  is  3,  inconsistent. 

4.  A  and  B  of  rank  2,  x  =3,  y  =2. 

Pages  126,  127 

1.  x=— — — — 1 r,  if  a,  b,  c  are  distinct  and  not  zero  and  their 

a(b  —a)(c  —a)  (a  +b  +c) 

sum  5^0.    If  a  =b?^c,  ac^O,  equations  are  inconsistent  unless  k  =0,  a,  c,  or  —  a  —  c, 

,  ,  k(c-k)  k(k-a)  ,.-, 

and  then  y  =— r  -x,   z  =  — r,  a;  arbitrary. 

a(c-a)  c(c—a) 

3.  (o-6)(6-c)(c-a).  4-  (x-y)(y-z)(z-x)(xy+yz+zx). 

6.  (a+6+c+d)  (a+6-c-d)  (0-6-c+d)  (a-b+c-d): 

7.  (a+6+c+d)  (a—  &+c— d)  (a+bi-c-di)  (a-bi-c+di). 

n 

11.  xj  =  {k1-aj)  .  .  .  (&« — %)-^-II  (as—aj). 

s=  1 

12.  x(a6+ac+6c)  =  —  ata. 


ANSWERS 


165 


Pages  133,  134 


p4  —  3p2q+5pr+q- 


r—pq 
5-  2p2-2q. 

3p2q2  —4q>*r  -4q3  -2pqr  -9r2 
(r-pq)2 


10.  y=q+r/x. 

4x2+px+q 


(5p2-12q)(p2-4q)  _13 
4(p3-4pq+8r)     ~~ZP' 
6.  24r-p3. 

8.  27r2  -9pqr  +2q*  =  0. 


11.  x  = 


12.  ?/=- 


— 3x-p 


,  see  §112. 


13. 


2+2y- 
2q(p3+2pq—r) 
p2q—pr+s 


-5p,  see  Ex.  17. 


Page  136 


3.  s2=p2-2q,  s3=p3-3pq,  st=p4-4:p2q+2q2,   s5=p5-5  p3q+5  pq2. 

4.  S5„=5-3n,  st=0  if  k  is  not  divisible  by  5.  5.  All  zero. 

Page  140 

2.  See  Ex.  2,  p.  136. 

3.  e^ic  +  VQ  +  e5--'^-  VQ,    Q  =  fc2-?6  (j=0,  1,  2,  3,  4). 

4.  e^c  +  VQ  +  e7-^-  VQ,    Q  =  \c2-qi  (j=0,  1,  .  .  .  ,  6). 


1.  c22—2cic3+2ci. 
3.  dc3-4c4. 


Page  141 

2.  C!2c2—  2c22—  CiC3+4c4. 
4.  c32— 2c2c4. 


Page  142 

1.  C!C3— 4c4 if  n> 3,  C1C3 if  n  =  3.  2.  3ciC4— c2c3— 5c6. 

3.  c2c4-4cic5+9c6.  4.  c32-2c2c4+2cic5-2c6. 

5.  y3  -(p2  -2q)y2  +  (q2 -2pr)y  -r2  =  0. 

6.  y3-qy2+pry-r2=0.  7.  n/3+2gt/2+4pz/+8  =0. 
8.  Eliminate  x  by  y=  s2- x2.  9.  Use  p2-q+px=y. 

10.    -4+pr/s.  11.   (rs-pr2+2pgs)/s2. 

12.    (i)  sasbscsd-'2sasbSc+d+2'2saSb+c+d  +  '2sa-H)sc+a-6sa+b+e+d. 
(u)  -^(V  -6sa2s2a  +8sas3a  +3s2a2  -6s4a). 


166 


ANSWERS 


13.    (0 


st  = 


1 

0 

0 

.  .     0 

Ci 

Ci 

1 

0 

.  .     0 

2c2 

c2 

Ci 

1 

.  .    0 

3c3 

C3 

C2 

Ci 

.  .    0 

4c4 

Ci- 

1     Cfc-2 

Cfc-3 

.  .  .    d 

&Ci 

s3=  - 


1       0        ci 

Ci       1        2c2 
c2       Ci       3c3 


where  all  but  the  last  term  in  the  main  diagonal  is  1,  and  all  terms  above  the 
diagonal  are  zero  except  those  in  the  last  column.  If  k>n,  we  must  take 
Cj=0  (j>n). 


(u) 


k !  Cjt  =  — 


1        0 

0 

.    0 

Si 

Si       2 

0 

.    0 

s2 

S2          Si 

3 

.    0 

s3 

S1-1    Sj— 2 

Si- 3     • 

•      Si 

s* 

,   3!c3=- 


1 

0 

Si 

Si 

2 

s2 

s2 

Si 

S3 

Page  152 

1.  y2{l%-y2);  y=0,x  =  ±3;  y  =  ±4,x  =  +5. 

2.  (c-l)%2-25)(?/2-16).    Ifc^l,    ?/  =  ±5,x=0;  y  =  ±4,a;  =  +3. 

3- 

=46 -a2.  4.  2±3i,  -2±i,  ±t. 

Pages  153,  154 

2.  pqr—p2s—r2=0,  x2-\-r/p,  x2+px+ps/r. 

3.  1,  3,  l±i.  11.  See  Ex.  15,  p.  134 


1 

a 

b 

2 

a 

0 

0 

2 

a 

INDEX 

Numbers  refer  to  pages. 


Abscissa,  55 

Absolute  value,  3 

Amplitude,  3 

Argument,  3 

Arithmetical  progression,  19 

Bend  point,  56,  64 
Bezout's  eliminant,  148 
Budan's  theorem,  83 

Cardan's  formulas,  46,  48 
Complex  number,  1 

geometrical  representation,  3,  6,  7 

trigonometric  form,  3 
Compound  interest,  13,  90,  100 
Conjugate,  1 
Continuity,  66,  157 
Cube  root,  5,  48 

of  unity,  3,  4 
Cubic  equation,  32,  40,  45,  127,  134,  153-4 

graph  of,  65 

number  real  roots,  48,  65,  79 

reduced,  45,  64-5 

trigonometric  solution,  49 

De  Moivre's  quintic,  140 

theorem,  5 
Derivative,  57-60,  69,  83,  97,  135 
Descartes'  rule  of  signs,  71,  85 
Determinants,  101-27,  146-53 

addition  of  columns,  113 

columns,  103 

complementary  minors,  122 

diagonal  term,  103 

elements,  103 

expansion,  109 

interchanges,  106,  107 

Laplace's  development,  122-3,  147-9 

minors,  109,  116 

of  Vandermonde,  108 

product  of,  124 

rank,  116,  121 


Determinants,  removal  of  factor,  111 

rows,  103 

signs  of  terms,  103-6 

skew  symmetric,  108 

sum  of,  112 
Discriminant,  152-3 

of  cubic,  47,  65,  134 

of  quadratic,  11,  12 

of  quartic,  51,  81 
Double  root,  16  (see  Discriminant) 
Duplication  of  cube,  35 

Elementary  symmetric  function,  128 
Elimination,  143-154 

extraneous  factor,  151 
Equation  for  differences  of  roots,  154 

squares  of  differences,  134,  154 
Euler's  eliminant,  147 

Factor  theorem,  12 

Factored  form,  11,  15 

Fundamental  theorem  of  algebra,  17,  155- 

Geometrical  construction,  29-44 

progression,  13,  19 
Graphs,  55-70 
Greatest  common  divisor,  61,  75 

Horner's  method,  86 

Identical  polynomials,  16 
Identity,  11 
Imaginary,  2 

roots,  19,  98 
Inflexion,  62-64 
Integral  rational  function,  12,  17,  20 

roots,  24-27 
Interpolation,  93,  97 
Interval,  78 
Irreducible  case,  48 
Isolation  of  roots,  71 


167 


168 


INDEX 


Linear  equations,  system,  101-3,  114-21 

homogeneous,  119,  121 
Linear  factors,  11,  17 
Lower  limit  to  roots,  23 

Matrix,  120 

augmented,  121 
Maximum,  57 
Minimum,  57,  157 
Modulus,  3 

Multiple  roots,  16,  60,  82 
Multiplicity  of  root,  16,  20,  61 

Newton's  identities,  136 

method  of  solution,  90-98 
Number  of  roots,  16,  17,  48,  52,  69,  72-85 

of  negative  roots,  74 

Order  of  radical,  32 
Ordinate,  55 

Plotting,  55 
Polynomial,  12,  66 

sign  of,  68 
Primitive  root  of  unity,  9 
Product  of  roots,  18 
Pure  imaginary,  2 

Quadratic  equation,  11 

graphical  solution,  29,  55 
sum  of  powers  of  roots,  139 

Quadratic  function  a  square,  12 

Quartic  equation,  50-54,  80-81 

Quotient  by  synthetic  division,  14 

Rational  roots,  27 
Real  equation,  12,  20 
Reciprocal  equation,  37,  44 
Regula  falsi,  93 
Regular  polygon,  8 

7  sides,  35-36 

9  sides,  35,  39 

17  sides,  41-44 

n  sides,  44 


Regular  decagon,  39 

pentagon,  39 
Relations  between  roots  and  coefficients,  17 
Relatively  prime,  9,  10 
Remainder  theorem,  12 
Resolvent  cubic,  50,  51 
Resultant,  143-154 
Rolle's  theorem,  69 
Root  between  a  and  b,  67 
Roots  of  unity,  8,  36,  39,  44,  136 

periods  of,  40 
Roots,  nth,  7 

Sigma  function,  128-142 

Sign  of  polynomial,  68 

Simple  root,  16 

Slope,  57,  59 

Solution  of  numerical  equations,  86-100 

Specific  gravity,  89 

Square  roots,  1,  30,  31,  96 

Sturm's  functions,  75-82 

Sum  of  four  squares,  126 

like  powers  of  roots,  134-142 

products  of  roots,  18 

roots,  18 
Surd  roots  in  pairs,  20 
Sylvester's  eliminant,  145,  149,  150 
Symbols,  11;  f(x),  12;    \a\,  23,  155;   r!,  59' 
/W(x),59;  S,  128;    sk,   134;  R(f,  g) 
144 
Symmetric  functions,  128-142 

in  all  but  one  root,  132-4 
Synthetic  division,  13,  86-95 

Tangents,  60,  62 
Taylor's  theorem,  59 
Transformed  equation,  28,  86 
Triple  root,  16 
Trisection  of  angle,  34,  40 

Upper  limit  to  roots,  21-23 

Variation  of  sign,  71 

Waring's  formula,  136 


BOSTON  COLLEGE 


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Catherine  B.  O'Connor  Library 

Weston  Observatory 

Weston  93,  Massachusetts 


